Strategies For Solving Trigonometric Equations

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Strategies For Solving Trigonometric Equations

An equation involving one or more trigonometrical ratios of unknown angles is called a trigonometrical equation.

1.0Solutions Of Trigonometric Equations

A value of the unknown angle which satisfies the given equation is called a solution of the trigonometric equation.

(a) Principal solution – The solution of the trigonometric equation lying in the interval [0, 2π).

(b) General solution – Since all the trigonometric functions are many one & periodic, hence there are infinite values of θ for which trigonometric functions have the same value. All such possible values of θ for which the given trigonometric function is satisfied is given by a general formula. Such a general formula is called general solution of trigonometric equation.

(c) Particular solution – The solution of the trigonometric equation lying in the given interval.

S. No.	Equation	Interval in which principal solution α	General solution
1.	sinθ = k, (–1 ≤ k ≤ 1)	$[- \frac{π}{2}, \frac{π}{2}]$	θ = nπ + (–1)ⁿ α, n∈Z
2.	cosθ = k, (–1 ≤ k ≤ 1)	[0, π]	θ = 2nπ ± α, n∈Z
3.	tanθ = k, (k∈R)	$(- \frac{π}{2}, \frac{π}{2})$	θ = nπ + α, n∈Z
4.	cosecθ = k, k ∈(–∞,–1]∪[1, ∞)	$[- \frac{π}{2}, \frac{π}{2}] - {0}$	θ = nπ + (–1)ⁿ α, n∈Z
5.	secθ = k, k ∈(–∞,–1]∪[1, ∞)	$[0, π] ∖ {\frac{π}{2}}$	θ = 2nπ ± α, n∈Z
6.	cotθ = k, k ∈ R	(0, π)	θ = nπ + α, n∈Z

2.0General Solutions of Some Trigonometric Equations

(a) If sin θ = 0, then θ = nπ, n ∈ I (set of integers)

(b) If cos θ = 0, then θ = (2n+1) $\frac{π}{2}$ , n ∈ I

(c) If tan θ = 0, then θ = nπ, n ∈ I

(d) If sin θ = sin α, then θ = nπ + (–1)ⁿα ; where $α \in [- \frac{π}{2}, \frac{π}{2}]$ , n ∈ I

(e) If cos θ = cos α, then θ = 2nπ ± α ; where n ∈ I and α ∈ [0,π]

(f) If tan θ = tan α, then θ = nπ + α ; where n ∈ I and $α \in [- \frac{π}{2}, \frac{π}{2}]$

(g) If sin θ =1, then θ = 2nπ + $\frac{π}{2}$ = (4n + 1) $\frac{π}{2}$ ; where n ∈ I

(h) If cos θ = 1 then θ = 2nπ ; where n ∈ I

(i) If sin² θ = sin² α or cos² θ = cos² α or tan² θ = tan² α

then θ = nπ ± α ; where n ∈ I

(j) For n ∈ I, sin nπ = 0 and cos nπ = (–1)ⁿ

sin (nπ + θ) = (–1)ⁿ sin θ

cos (nπ + θ) = (–1)ⁿ cos θ

(k) If n is an odd integer, then $sin \frac{nπ}{2} = (- 1)^{\frac{n - 1}{2}}, cos \frac{nπ}{2} = 0,$ $sin (\frac{nπ}{2} + θ) = (- 1)^{\frac{n - 1}{2}} cos θ,$ $cos (\frac{nπ}{2} + θ) = (- 1)^{\frac{n + 1}{2}} sin θ$

3.0Techniques for Simplifying and Solving Trigonometric Equations

Solving Trigonometric Equations by Factorisation

e.g. (2 sin x – cos x) (1 + cos x) = sin²x

⇒ (2 sin x – cos x) (1 + cos x) – (1 – cos²x) = 0

⇒ (1 + cos x) (2 sin x – cos x – 1 + cos x) = 0

⇒ (1 + cos x) (2 sin x – 1) = 0

⇒ cos x = –1 or sin x = $\frac{1}{2}$

⇒ cosx = – 1 = cosπ

⇒ x = 2nπ ± π = (2n ± 1)π, n ∈ I

or sinx = $\frac{1}{2} = s in \frac{π}{6}$

⇒ x = kπ + (–1)^k $\frac{π}{6}$ , k ∈ I

Solving Trigonometric Equations by Introducing an Auxiliary Argument

Consider, a sin θ + b cos θ = c .....(i)

∴ $\frac{a}{a ^{2} + b ^{2}} sin θ + \frac{b}{a ^{2} + b ^{2}} cos θ = \frac{c}{a ^{2} + b ^{2}}$

equation (i) has a solution only if |c| $\leq a^{2} + b^{2}$

let $\frac{a}{a ^{2} + b ^{2}} = cos ϕ, \frac{b}{a ^{2} + b ^{2}} = sin ϕ & ϕ = tan^{- 1} \frac{b}{a}$ by introducing this auxiliary argument φ, equation (i) reduces to

sin (θ + φ) = $sin (θ + ϕ) = \frac{c}{a ^{2} + b ^{2}}$ , Now this equation can be solved easily.

4.0Solved Examples

1. Find the general value of x satisfying the equation $3 sin x + cos x = 3$

Solution

We have $3 sin x + cos x = 3$

Dividing both sides by $a^{2} + b^{2}$ i.e. , $4 = 2$ we get

$\frac{1}{2} cos x + \frac{3}{2} sin x = \frac{3}{2}$

$\Rightarrow cos x cos \frac{π}{3} + sin x sin \frac{π}{3} = \frac{3}{2}$

$\Rightarrow cos (x - \frac{π}{3}) = cos \frac{π}{6}$

$\Rightarrow x - \frac{π}{3} = 2 nπ \pm \frac{π}{6}, n \in I$

$x - \frac{π}{2} = 2 nπ + \frac{π}{6}, x - \frac{π}{2} = 2 nπ - \frac{π}{6}; n \in I$

Solving Trigonometric Equation by Reducing it to a Quadratic Equation

e.g. 6 – 10cosx = 3sin²x

⇒ 6 – 10cosx = 3 – 3cos²x

⇒ 3cos²x – 10cosx + 3 = 0

⇒ 3cosx – 1) (cosx – 3) = 0

⇒ cosx = $\frac{1}{3}$ or cosx = 3

cosx = 3 is not possible as – 1 ≤ cosx ≤ 1

∴ cosx = $\frac{1}{3}$ = cos $(cos^{- 1} \frac{1}{3})$

⇒ x = 2nπ ± cos^–1 $(\frac{1}{3}), n \in I$

Solved Examples

1. Solve sin²θ − cosθ = $\frac{1}{4}$ for θ and write the values of θ in the interval 0 ≤ θ ≤ 2π.

Solution

The given equation can be written as

1 – cos²θ – cosθ = $\frac{1}{4}$

⇒ cos²θ + cosθ – 3/4 = 0

⇒ 4cos²θ + 4cosθ – 3 = 0

⇒ (2cosθ – 1)(2cosθ + 3) = 0

⇒ cosθ = $\frac{1}{2}, \frac{3}{- 2}$

Since, cosθ = –3/2 is not possible as –1 ≤ cosθ ≤ 1

∴ $cos θ = \frac{1}{2}$

$cos θ = cos \frac{π}{3} ⟹ θ = 2 nπ \pm \frac{π}{3}, n \in I$

For the given interval, n = 0 and n = 1.

$θ = \frac{π}{3}, \frac{5 π}{3}$

2. Find the number of solutions of tanx + secx = 2cosx in [0, 2π].

Solution

Here, tanx + secx = 2cosx

⇒ sinx + 1 = 2 cos²x

⇒ 2sin²x + sinx – 1 = 0

⇒ sinx = $\frac{1}{2}$ , – 1

But sinx = –1 ⇒ x = $\frac{3 π}{2}$ for which tanx + secx = 2 cosx is not defined.

Thus sinx = $\frac{1}{2}$ ⇒ x = $\frac{π}{6}, \frac{5 π}{6}$

⇒ number of solutions of tanx + secx = 2cos x is 2.

3. Solve the equation 5sin²x – 7sinx cosx + 16cos² x = 4

Solution

To solve this equation we use the fundamental formula of trigonometric identities,

sin²x + cos²x = 1

writing the equation in the form,

5sin²x – 7sinx . cosx + 16cos²x = 4(sin²x + cos²x)

⇒ sin²x – 7sinx cosx + 12cos² x = 0

dividing by cos²x on both side we get,

tan²x – 7tanx + 12 = 0

Now it can be factorized as :

(tanx – 3)(tanx – 4) = 0

⇒ tanx = 3, 4 ⇒ x = nπ + tan^–1 3

or x = mπ + tan^–1 4, m ∈ I.

4. If x ≠ $\frac{nπ}{2}$ , n ∈ I and $(cos x) sin^{2} x - 3 sin x + 2 = 1$ , then find the general solutions of x.

Solution

As $\frac{nπ}{2}$ ⇒ cos x ≠ 0, 1, – 1

So, $(cos x) sin^{2} x - 3 sin x + 2 = 1$

⇒ sin²x – 3sinx + 2 = 0 ⇒ (sinx – 2) (sinx – 1) = 0

⇒ sinx = 1, 2

where sinx = 2 is not possible and sinx = 1 which is also not possible as $x \neq = \frac{nπ}{2}$

∴ No solution is possible.

5. Solve the equation sin⁴x + cos⁴ $\frac{7}{2} s in x . cos x .$

Solution

sin⁴x + cos⁴x = $\frac{7}{2} s in x . cos x$

⇒ (sin²x + cos²x)² – 2sin²x cos² $\frac{7}{2} s in x . cos x$

⇒ $1 - \frac{1}{2} (sin 2 x)^{2} = \frac{7}{4} (sin 2 x)$ ⇒ 2sin²2x + 7sin2x – 4 = 0

⇒ (2sin2x –1)(sin2x + 4) = 0

⇒ sin2x = $\frac{1}{2}$ or sin2x = –4 (which is not possible)

⇒ 2x = nπ + (–1)ⁿ $\frac{π}{6}$ , n ∈ I

i.e., $x = \frac{nπ}{2} + (- 1)^{n} \frac{π}{12}, n \in I$

Solving Trigonometric Equations by Introducing an Auxiliary Argument

Consider, a sin θ + b cos θ = c .....(i)

∴ $\frac{a}{a ^{2} + b ^{2}} sin θ + \frac{b}{a ^{2} + b ^{2}} cos θ = \frac{c}{a ^{2} + b ^{2}}$

equation (i) has a solution only if |c| ≤ $a^{2} + b^{2}$

let $\frac{a}{a ^{2} + b ^{2}} = cos ϕ, \frac{b}{a ^{2} + b ^{2}} = sin ϕ and ϕ = tan^{- 1} \frac{b}{a}$ by introducing this auxiliary argument φ, equation (i) reduces to sin (θ + φ) = $\frac{c}{a ^{2} + b ^{2}}$ , Now this equation can be solved easily.

6. Find the number of distinct solutions of secx + tanx = $sec x + tan x = 3$ , where 0 ≤ x ≤ 3π.

Solution

Here, sec x + tanx = $3$

⇒ 1 + sinx = $3$ cosx

or $3$ cosx – sinx = 1

dividing both sides by $a^{2} + b^{2} i . e . 4 = 2$ , we get $\frac{3}{2} cos x - \frac{1}{2} sin x = \frac{1}{2}$

$\Rightarrow cos \frac{π}{6} cos x - sin \frac{π}{6} sin x = \frac{1}{2}$

$\Rightarrow cos (x + \frac{π}{6}) = \frac{1}{2}$

As 0 ≤ x ≤ 3π

$\frac{π}{6} \leq x + \frac{π}{6} \leq 3 π + \frac{π}{6}$

$x + \frac{π}{6} = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3}$

$x = \frac{π}{6}, \frac{3 π}{2}, \frac{13 π}{6}$

But at x= $\frac{3 π}{2}$ ; tanx and secx is not defined.

∴ Total number of solutions are 2.

7. Find the general value of x satisfying the equation $3 sin x + cos x = 3$

Solution

We have $3 sin x + cos x = 3$

Dividing both sides by $a^{2} + b^{2} i . e . 4 = 2,$ we get $\frac{1}{2} cos x + \frac{3}{2} sin x = \frac{3}{2} .$

$\Rightarrow cos x cos \frac{π}{3} + sin x sin \frac{π}{3} = \frac{3}{2}$

$\Rightarrow cos (x - \frac{π}{3}) = cos \frac{π}{6}$

$\Rightarrow x - \frac{π}{3} = 2 nπ \pm \frac{π}{6}, n \in I$

$\Rightarrow x - \frac{π}{3} = 2 nπ + \frac{π}{6}, + x - \frac{π}{3} = 2 nπ + \frac{π}{6} n \in Z .$

$\Rightarrow x = 2 nπ + \frac{π}{2}, x = 2 nπ + \frac{π}{2}; n \in Z .$

5.0Techniques Of Trigonometric Equations Solution

Solving Trigonometric Equations By Transforming Sum Of Trigonometric Functions Into Product

e.g. cos 3x + sin 2x – sin 4x = 0

cos 3x – 2 sin x cos 3x = 0

⇒ (cos3x) (1 – 2sinx) = 0

⇒ cos3x = 0 or sinx = $\frac{1}{2}$

⇒ cos3x = 0= cos $\frac{π}{2}$ or sinx = $\frac{1}{2}$ = sin $\frac{π}{6}$

⇒ 3x = 2nπ ± $\frac{π}{2}$ or x = mπ + (–1)^m $\frac{π}{6}$

⇒ x = $\frac{2 nπ}{3} \pm \frac{π}{6}$

or x = mπ + (–1)^m $\frac{π}{6}$ ; (n, m ∈ I)

Solving trigonometric equations by a change of variable

(i) Equations of the form P (sin x ± cos x) = 0, P(sin x. cos x) = 0 can be solved by the substitution, cos x ± sin x = t ⇒ 1 ± 2 sin x. cos x = t².

(ii) Equations of the form of asinx + bcosx + d = 0, where a, b & d are real numbers can be solved by changing sin x & cos x into their corresponding tangent of half the angle.

(iii) Many equations can be solved by introducing a new variable.

Solved Examples

1. Solve : cos x + cos 2x + cos 3x = 0.

Solution

cos x + cos 2x + cos 3x = 0

We shall rearrange and use the transformation formula cos 2x + (cos x + cos 3x) = 0

By using the formula, $cos A + cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2})$

⇒ cos 2x + $2 cos (\frac{3 x + x}{2}) cos (\frac{3 x - x}{2}) = 0$

⇒ cos 2x + 2cos 2x cos x = 0

⇒ cos 2x ( 1 + 2 cos x) = 0

⇒ cos 2x = 0 or 1 + 2cos x = 0

⇒ cos 2x = 0 or cos x = $- \frac{1}{2}$

⇒ $cos 2 x = cos \frac{π}{2} or cos x = cos (π - \frac{π}{3})$

⇒ $cos 2 x = cos \frac{π}{2} or cos x = cos (\frac{2 π}{3})$

⇒ $2 x = (2 n + 1) \frac{π}{2} or x = 2 mπ \pm \frac{2 π}{3}$

⇒ $x = (2 n + 1) \frac{π}{4} or x = 2 mπ \pm \frac{2 π}{3}$

∴ the general solution is

x = (2n + 1) $\frac{π}{4}$ or 2mπ ± $\frac{2 π}{3}$ , where m, n ϵ I

e.g. sin5x. cos3x = sin6x. cos2x

sin8x + sin2x = sin8x + sin4x

∴ 2sin2x . cos2x – sin2x = 0

⇒ sin2x(2 cos 2x – 1) = 0

sin2x = 0 or 2 cos 2x – 1= 0

sin2x = 0 = sin0

⇒ 2x = nπ+(–1)ⁿ × 0, n ∈ I

⇒ x = $\frac{nπ}{2}$ , n ∈ I or

2 cos 2x – 1= 0

⇒ cos2x = $\frac{1}{2}$

⇒ cos2x = $\frac{1}{2}$ = cos $\frac{π}{3}$

⇒ 2x = 2mπ ± $\frac{π}{3}$ , m ∈ I

⇒ x = mπ ± $\frac{π}{6}$ , m ∈ I

2. Find the number of solutions in $[0, \frac{π}{2}]$ of the equation cos3x tan5x = sin7x.

Solution

We have cos3x tan5x = sin7x

⇒ $cos 3 x (\frac{s i n 5 x}{c o s 5 x}) = sin 7 x$

⇒ 2cos3x sin5x = 2cos 5x sin7x

⇒ sin8x + sin2x = sin12x + sin2x

⇒ sin12x – sin8x = 0

⇒ 2sin2x cos10x = 0

If sin 2x = 0, then $x \in [0, \frac{π}{2}]$

If cos 10x = 0, then

$10 x = \frac{π}{2}, \frac{3 π}{2}$

$\Rightarrow x = \frac{π}{20}, \frac{3 π}{20}, \frac{5 π}{20}, \frac{7 π}{20}, \frac{9 π}{20}$

So, there are 6 possible solutions.

3. Solve : cosθ + cos3θ + cos5θ + cos7θ = 0

Solution

We have cosθ + cos7θ + cos3θ + cos5θ = 0

⇒ 2cos4θcos3θ + 2cos4θcosθ = 0

⇒ cos4θ(cos3θ + cosθ) = 0

⇒ cos4θ(2cos2θcosθ) = 0

⇒ cosθ = 0 ⇒ θ = (2n₁ + 1) $(\frac{π}{2})$ , n₁ ∈ I

or cos2θ = 0 ⇒ θ = (2n₂ + 1) $(\frac{π}{4})$ , n₂ ∈ I

or cos4θ = 0 ⇒ θ = (2n₃ + 1) $(\frac{π}{8})$ , n₃ ∈ I

Solving trigonometric equations with the use of the boundness of the functions involved

4. Solve : $sin x (cos \frac{x}{4} - 2 sin x) + (1 + sin \frac{x}{4} - 2 cos x) \cdot cos x = 0.$

Solution

$sin x (cos \frac{x}{4} - 2 sin x) + (1 + sin \frac{x}{4} - 2 cos x) \cdot cos x = 0.$

∴ sin x cos $\frac{x}{4} + cos x sin \frac{x}{4} + cos x = 2$

∴ sin $(\frac{5 x}{4}) + cos x = 2$

⇒ sin $(\frac{5 x}{4}) = 1$ & cos x =1 (as sin θ ≤ 1 & cos θ ≤ 1)

Now consider

cosx = 1

⇒ x = 2π, 4π, 6π, 8π .......

And sin $\frac{5 x}{4} = 1$

⇒ x = $\frac{2 π}{5}, \frac{10 π}{5}, \frac{18 π}{5}$ ......

Common solution to above APs will be the AP having

First term = 2π

Common difference = LCM of 2π and $\frac{8 π}{5} = \frac{40 π}{5} = 8 π$

∴ General solution will be general term of this AP i.e. 2π + (8π)n, n ∈ I

⇒ x = 2(4n + 1)π, n ∈ I

5. Solve the equation (sinx + cosx)^1+sin2x= 2, when $0 \leq x \leq π$

Solution

We know, $- a^{2} + b^{2} \leq a sin θ + b cos θ \leq a^{2} + b^{2}$ and –1 ≤ sinθ ≤ 1.

∴ (sinx + cosx) admits the maximum value as $2$ and (1 + sin 2x) admits the maximum value as 2.

Also $(2)^{2} = 2$

∴ the equation could hold only when, sinx + cosx = $2$ and 1 + sin 2x = 2

Now, sinx + cos x = $2$

⇒ $cos (x - \frac{π}{4}) = 1$

⇒ x = 2nπ + $\frac{π}{4}$ , n ∈ I ...... (i)

and 1 + sin 2x = 2

⇒ sin2x = 1 = sin $\frac{π}{2}$

⇒ 2x = mπ + (–1)^m $\frac{π}{2}$ , m ∈ I

⇒ $x = \frac{mπ}{2} + (- 1)^{m} \frac{π}{4}$ ...... (ii)

The value of x in [0, π] satisfying equations (i) and (ii) is x = $\frac{π}{4}$

(when n = 0 & m = 0)

Note :- sin x + cos x = $- 2$ and 1 + sin 2x = 2 also satisfies but as x > 0, this solution is not in domain.

6. Find the set of values of x for which $\frac{t a n 3 x - t a n 2 x}{1 + t a n 3 x \cdot t a n 2 x} = 1.$

Solution

We have, $\frac{t a n 3 x - t a n 2 x}{1 + t a n 3 x \cdot t a n 2 x} = 1.$

⇒ tan (3x – 2x) = 1

⇒ tan x = 1

⇒ tan x = tan $\frac{π}{4}$

⇒ x = nπ + $\frac{π}{4}$ , n ∈ I {using tanθ = tanα ⇔ θ = nπ + α}

But for this value of x, tan 2x is not defined. Hence the solution set for x is φ.

6.0Also Read

Standard Units of Measurement	Angles	Introduction to Numbers
Ascending Order	Horizontal Line	Roman Numerals
Area of Irregular Shapes	Laws of Exponents	Rounding Numbers

Strategies For Solving Trigonometric Equations

An equation involving one or more trigonometrical ratios of unknown angles is called a trigonometrical equation.

1.0Solutions Of Trigonometric Equations

A value of the unknown angle which satisfies the given equation is called a solution of the trigonometric equation.

(a) Principal solution – The solution of the trigonometric equation lying in the interval [0, 2π).

(b) General solution – Since all the trigonometric functions are many one & periodic, hence there are infinite values of θ for which trigonometric functions have the same value. All such possible values of θ for which the given trigonometric function is satisfied is given by a general formula. Such a general formula is called general solution of trigonometric equation.

(c) Particular solution – The solution of the trigonometric equation lying in the given interval.

S. No.	Equation	Interval in which principal solution α	General solution
1.	sinθ = k, (–1 ≤ k ≤ 1)	$[- \frac{π}{2}, \frac{π}{2}]$	θ = nπ + (–1)ⁿ α, n∈Z
2.	cosθ = k, (–1 ≤ k ≤ 1)	[0, π]	θ = 2nπ ± α, n∈Z
3.	tanθ = k, (k∈R)	$(- \frac{π}{2}, \frac{π}{2})$	θ = nπ + α, n∈Z
4.	cosecθ = k, k ∈(–∞,–1]∪[1, ∞)	$[- \frac{π}{2}, \frac{π}{2}] - {0}$	θ = nπ + (–1)ⁿ α, n∈Z
5.	secθ = k, k ∈(–∞,–1]∪[1, ∞)	$[0, π] ∖ {\frac{π}{2}}$	θ = 2nπ ± α, n∈Z
6.	cotθ = k, k ∈ R	(0, π)	θ = nπ + α, n∈Z

2.0General Solutions of Some Trigonometric Equations

(a) If sin θ = 0, then θ = nπ, n ∈ I (set of integers)

(b) If cos θ = 0, then θ = (2n+1) $\frac{π}{2}$ , n ∈ I

(c) If tan θ = 0, then θ = nπ, n ∈ I

(d) If sin θ = sin α, then θ = nπ + (–1)ⁿα ; where $α \in [- \frac{π}{2}, \frac{π}{2}]$ , n ∈ I

(e) If cos θ = cos α, then θ = 2nπ ± α ; where n ∈ I and α ∈ [0,π]

(f) If tan θ = tan α, then θ = nπ + α ; where n ∈ I and $α \in [- \frac{π}{2}, \frac{π}{2}]$

(g) If sin θ =1, then θ = 2nπ + $\frac{π}{2}$ = (4n + 1) $\frac{π}{2}$ ; where n ∈ I

(h) If cos θ = 1 then θ = 2nπ ; where n ∈ I

(i) If sin² θ = sin² α or cos² θ = cos² α or tan² θ = tan² α

then θ = nπ ± α ; where n ∈ I

(j) For n ∈ I, sin nπ = 0 and cos nπ = (–1)ⁿ

sin (nπ + θ) = (–1)ⁿ sin θ

cos (nπ + θ) = (–1)ⁿ cos θ

(k) If n is an odd integer, then $sin \frac{nπ}{2} = (- 1)^{\frac{n - 1}{2}}, cos \frac{nπ}{2} = 0,$ $sin (\frac{nπ}{2} + θ) = (- 1)^{\frac{n - 1}{2}} cos θ,$ $cos (\frac{nπ}{2} + θ) = (- 1)^{\frac{n + 1}{2}} sin θ$

3.0Techniques for Simplifying and Solving Trigonometric Equations

Solving Trigonometric Equations by Factorisation

e.g. (2 sin x – cos x) (1 + cos x) = sin²x

⇒ (2 sin x – cos x) (1 + cos x) – (1 – cos²x) = 0

⇒ (1 + cos x) (2 sin x – cos x – 1 + cos x) = 0

⇒ (1 + cos x) (2 sin x – 1) = 0

⇒ cos x = –1 or sin x = $\frac{1}{2}$

⇒ cosx = – 1 = cosπ

⇒ x = 2nπ ± π = (2n ± 1)π, n ∈ I

or sinx = $\frac{1}{2} = s in \frac{π}{6}$

⇒ x = kπ + (–1)^k $\frac{π}{6}$ , k ∈ I

Solving Trigonometric Equations by Introducing an Auxiliary Argument

Consider, a sin θ + b cos θ = c .....(i)

∴ $\frac{a}{a ^{2} + b ^{2}} sin θ + \frac{b}{a ^{2} + b ^{2}} cos θ = \frac{c}{a ^{2} + b ^{2}}$

equation (i) has a solution only if |c| $\leq a^{2} + b^{2}$

let $\frac{a}{a ^{2} + b ^{2}} = cos ϕ, \frac{b}{a ^{2} + b ^{2}} = sin ϕ & ϕ = tan^{- 1} \frac{b}{a}$ by introducing this auxiliary argument φ, equation (i) reduces to

sin (θ + φ) = $sin (θ + ϕ) = \frac{c}{a ^{2} + b ^{2}}$ , Now this equation can be solved easily.

4.0Solved Examples

1. Find the general value of x satisfying the equation $3 sin x + cos x = 3$

Solution

We have $3 sin x + cos x = 3$

Dividing both sides by $a^{2} + b^{2}$ i.e. , $4 = 2$ we get

$\frac{1}{2} cos x + \frac{3}{2} sin x = \frac{3}{2}$

$\Rightarrow cos x cos \frac{π}{3} + sin x sin \frac{π}{3} = \frac{3}{2}$

$\Rightarrow cos (x - \frac{π}{3}) = cos \frac{π}{6}$

$\Rightarrow x - \frac{π}{3} = 2 nπ \pm \frac{π}{6}, n \in I$

$x - \frac{π}{2} = 2 nπ + \frac{π}{6}, x - \frac{π}{2} = 2 nπ - \frac{π}{6}; n \in I$

Solving Trigonometric Equation by Reducing it to a Quadratic Equation

e.g. 6 – 10cosx = 3sin²x

⇒ 6 – 10cosx = 3 – 3cos²x

⇒ 3cos²x – 10cosx + 3 = 0

⇒ 3cosx – 1) (cosx – 3) = 0

⇒ cosx = $\frac{1}{3}$ or cosx = 3

cosx = 3 is not possible as – 1 ≤ cosx ≤ 1

∴ cosx = $\frac{1}{3}$ = cos $(cos^{- 1} \frac{1}{3})$

⇒ x = 2nπ ± cos^–1 $(\frac{1}{3}), n \in I$

Solved Examples

1. Solve sin²θ − cosθ = $\frac{1}{4}$ for θ and write the values of θ in the interval 0 ≤ θ ≤ 2π.

Solution

The given equation can be written as

1 – cos²θ – cosθ = $\frac{1}{4}$

⇒ cos²θ + cosθ – 3/4 = 0

⇒ 4cos²θ + 4cosθ – 3 = 0

⇒ (2cosθ – 1)(2cosθ + 3) = 0

⇒ cosθ = $\frac{1}{2}, \frac{3}{- 2}$

Since, cosθ = –3/2 is not possible as –1 ≤ cosθ ≤ 1

∴ $cos θ = \frac{1}{2}$

$cos θ = cos \frac{π}{3} ⟹ θ = 2 nπ \pm \frac{π}{3}, n \in I$

For the given interval, n = 0 and n = 1.

$θ = \frac{π}{3}, \frac{5 π}{3}$

2. Find the number of solutions of tanx + secx = 2cosx in [0, 2π].

Solution

Here, tanx + secx = 2cosx

⇒ sinx + 1 = 2 cos²x

⇒ 2sin²x + sinx – 1 = 0

⇒ sinx = $\frac{1}{2}$ , – 1

But sinx = –1 ⇒ x = $\frac{3 π}{2}$ for which tanx + secx = 2 cosx is not defined.

Thus sinx = $\frac{1}{2}$ ⇒ x = $\frac{π}{6}, \frac{5 π}{6}$

⇒ number of solutions of tanx + secx = 2cos x is 2.

3. Solve the equation 5sin²x – 7sinx cosx + 16cos² x = 4

Solution

To solve this equation we use the fundamental formula of trigonometric identities,

sin²x + cos²x = 1

writing the equation in the form,

5sin²x – 7sinx . cosx + 16cos²x = 4(sin²x + cos²x)

⇒ sin²x – 7sinx cosx + 12cos² x = 0

dividing by cos²x on both side we get,

tan²x – 7tanx + 12 = 0

Now it can be factorized as :

(tanx – 3)(tanx – 4) = 0

⇒ tanx = 3, 4 ⇒ x = nπ + tan^–1 3

or x = mπ + tan^–1 4, m ∈ I.

4. If x ≠ $\frac{nπ}{2}$ , n ∈ I and $(cos x) sin^{2} x - 3 sin x + 2 = 1$ , then find the general solutions of x.

Solution

As $\frac{nπ}{2}$ ⇒ cos x ≠ 0, 1, – 1

So, $(cos x) sin^{2} x - 3 sin x + 2 = 1$

⇒ sin²x – 3sinx + 2 = 0 ⇒ (sinx – 2) (sinx – 1) = 0

⇒ sinx = 1, 2

where sinx = 2 is not possible and sinx = 1 which is also not possible as $x \neq = \frac{nπ}{2}$

∴ No solution is possible.

5. Solve the equation sin⁴x + cos⁴ $\frac{7}{2} s in x . cos x .$

Solution

sin⁴x + cos⁴x = $\frac{7}{2} s in x . cos x$

⇒ (sin²x + cos²x)² – 2sin²x cos² $\frac{7}{2} s in x . cos x$

⇒ $1 - \frac{1}{2} (sin 2 x)^{2} = \frac{7}{4} (sin 2 x)$ ⇒ 2sin²2x + 7sin2x – 4 = 0

⇒ (2sin2x –1)(sin2x + 4) = 0

⇒ sin2x = $\frac{1}{2}$ or sin2x = –4 (which is not possible)

⇒ 2x = nπ + (–1)ⁿ $\frac{π}{6}$ , n ∈ I

i.e., $x = \frac{nπ}{2} + (- 1)^{n} \frac{π}{12}, n \in I$

Solving Trigonometric Equations by Introducing an Auxiliary Argument

Consider, a sin θ + b cos θ = c .....(i)

∴ $\frac{a}{a ^{2} + b ^{2}} sin θ + \frac{b}{a ^{2} + b ^{2}} cos θ = \frac{c}{a ^{2} + b ^{2}}$

equation (i) has a solution only if |c| ≤ $a^{2} + b^{2}$

let $\frac{a}{a ^{2} + b ^{2}} = cos ϕ, \frac{b}{a ^{2} + b ^{2}} = sin ϕ and ϕ = tan^{- 1} \frac{b}{a}$ by introducing this auxiliary argument φ, equation (i) reduces to sin (θ + φ) = $\frac{c}{a ^{2} + b ^{2}}$ , Now this equation can be solved easily.

6. Find the number of distinct solutions of secx + tanx = $sec x + tan x = 3$ , where 0 ≤ x ≤ 3π.

Solution

Here, sec x + tanx = $3$

⇒ 1 + sinx = $3$ cosx

or $3$ cosx – sinx = 1

dividing both sides by $a^{2} + b^{2} i . e . 4 = 2$ , we get $\frac{3}{2} cos x - \frac{1}{2} sin x = \frac{1}{2}$

$\Rightarrow cos \frac{π}{6} cos x - sin \frac{π}{6} sin x = \frac{1}{2}$

$\Rightarrow cos (x + \frac{π}{6}) = \frac{1}{2}$

As 0 ≤ x ≤ 3π

$\frac{π}{6} \leq x + \frac{π}{6} \leq 3 π + \frac{π}{6}$

$x + \frac{π}{6} = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3}$

$x = \frac{π}{6}, \frac{3 π}{2}, \frac{13 π}{6}$

But at x= $\frac{3 π}{2}$ ; tanx and secx is not defined.

∴ Total number of solutions are 2.

7. Find the general value of x satisfying the equation $3 sin x + cos x = 3$

Solution

We have $3 sin x + cos x = 3$

Dividing both sides by $a^{2} + b^{2} i . e . 4 = 2,$ we get $\frac{1}{2} cos x + \frac{3}{2} sin x = \frac{3}{2} .$

$\Rightarrow cos x cos \frac{π}{3} + sin x sin \frac{π}{3} = \frac{3}{2}$

$\Rightarrow cos (x - \frac{π}{3}) = cos \frac{π}{6}$

$\Rightarrow x - \frac{π}{3} = 2 nπ \pm \frac{π}{6}, n \in I$

$\Rightarrow x - \frac{π}{3} = 2 nπ + \frac{π}{6}, + x - \frac{π}{3} = 2 nπ + \frac{π}{6} n \in Z .$

$\Rightarrow x = 2 nπ + \frac{π}{2}, x = 2 nπ + \frac{π}{2}; n \in Z .$

5.0Techniques Of Trigonometric Equations Solution

Solving Trigonometric Equations By Transforming Sum Of Trigonometric Functions Into Product

e.g. cos 3x + sin 2x – sin 4x = 0

cos 3x – 2 sin x cos 3x = 0

⇒ (cos3x) (1 – 2sinx) = 0

⇒ cos3x = 0 or sinx = $\frac{1}{2}$

⇒ cos3x = 0= cos $\frac{π}{2}$ or sinx = $\frac{1}{2}$ = sin $\frac{π}{6}$

⇒ 3x = 2nπ ± $\frac{π}{2}$ or x = mπ + (–1)^m $\frac{π}{6}$

⇒ x = $\frac{2 nπ}{3} \pm \frac{π}{6}$

or x = mπ + (–1)^m $\frac{π}{6}$ ; (n, m ∈ I)

Solving trigonometric equations by a change of variable

(i) Equations of the form P (sin x ± cos x) = 0, P(sin x. cos x) = 0 can be solved by the substitution, cos x ± sin x = t ⇒ 1 ± 2 sin x. cos x = t².

(ii) Equations of the form of asinx + bcosx + d = 0, where a, b & d are real numbers can be solved by changing sin x & cos x into their corresponding tangent of half the angle.

(iii) Many equations can be solved by introducing a new variable.

Solved Examples

1. Solve : cos x + cos 2x + cos 3x = 0.

Solution

cos x + cos 2x + cos 3x = 0

We shall rearrange and use the transformation formula cos 2x + (cos x + cos 3x) = 0

By using the formula, $cos A + cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2})$

⇒ cos 2x + $2 cos (\frac{3 x + x}{2}) cos (\frac{3 x - x}{2}) = 0$

⇒ cos 2x + 2cos 2x cos x = 0

⇒ cos 2x ( 1 + 2 cos x) = 0

⇒ cos 2x = 0 or 1 + 2cos x = 0

⇒ cos 2x = 0 or cos x = $- \frac{1}{2}$

⇒ $cos 2 x = cos \frac{π}{2} or cos x = cos (π - \frac{π}{3})$

⇒ $cos 2 x = cos \frac{π}{2} or cos x = cos (\frac{2 π}{3})$

⇒ $2 x = (2 n + 1) \frac{π}{2} or x = 2 mπ \pm \frac{2 π}{3}$

⇒ $x = (2 n + 1) \frac{π}{4} or x = 2 mπ \pm \frac{2 π}{3}$

∴ the general solution is

x = (2n + 1) $\frac{π}{4}$ or 2mπ ± $\frac{2 π}{3}$ , where m, n ϵ I

e.g. sin5x. cos3x = sin6x. cos2x

sin8x + sin2x = sin8x + sin4x

∴ 2sin2x . cos2x – sin2x = 0

⇒ sin2x(2 cos 2x – 1) = 0

sin2x = 0 or 2 cos 2x – 1= 0

sin2x = 0 = sin0

⇒ 2x = nπ+(–1)ⁿ × 0, n ∈ I

⇒ x = $\frac{nπ}{2}$ , n ∈ I or

2 cos 2x – 1= 0

⇒ cos2x = $\frac{1}{2}$

⇒ cos2x = $\frac{1}{2}$ = cos $\frac{π}{3}$

⇒ 2x = 2mπ ± $\frac{π}{3}$ , m ∈ I

⇒ x = mπ ± $\frac{π}{6}$ , m ∈ I

2. Find the number of solutions in $[0, \frac{π}{2}]$ of the equation cos3x tan5x = sin7x.

Solution

We have cos3x tan5x = sin7x

⇒ $cos 3 x (\frac{s i n 5 x}{c o s 5 x}) = sin 7 x$

⇒ 2cos3x sin5x = 2cos 5x sin7x

⇒ sin8x + sin2x = sin12x + sin2x

⇒ sin12x – sin8x = 0

⇒ 2sin2x cos10x = 0

If sin 2x = 0, then $x \in [0, \frac{π}{2}]$

If cos 10x = 0, then

$10 x = \frac{π}{2}, \frac{3 π}{2}$

$\Rightarrow x = \frac{π}{20}, \frac{3 π}{20}, \frac{5 π}{20}, \frac{7 π}{20}, \frac{9 π}{20}$

So, there are 6 possible solutions.

3. Solve : cosθ + cos3θ + cos5θ + cos7θ = 0

Solution

We have cosθ + cos7θ + cos3θ + cos5θ = 0

⇒ 2cos4θcos3θ + 2cos4θcosθ = 0

⇒ cos4θ(cos3θ + cosθ) = 0

⇒ cos4θ(2cos2θcosθ) = 0

⇒ cosθ = 0 ⇒ θ = (2n₁ + 1) $(\frac{π}{2})$ , n₁ ∈ I

or cos2θ = 0 ⇒ θ = (2n₂ + 1) $(\frac{π}{4})$ , n₂ ∈ I

or cos4θ = 0 ⇒ θ = (2n₃ + 1) $(\frac{π}{8})$ , n₃ ∈ I

Solving trigonometric equations with the use of the boundness of the functions involved

4. Solve : $sin x (cos \frac{x}{4} - 2 sin x) + (1 + sin \frac{x}{4} - 2 cos x) \cdot cos x = 0.$

Solution

$sin x (cos \frac{x}{4} - 2 sin x) + (1 + sin \frac{x}{4} - 2 cos x) \cdot cos x = 0.$

∴ sin x cos $\frac{x}{4} + cos x sin \frac{x}{4} + cos x = 2$

∴ sin $(\frac{5 x}{4}) + cos x = 2$

⇒ sin $(\frac{5 x}{4}) = 1$ & cos x =1 (as sin θ ≤ 1 & cos θ ≤ 1)

Now consider

cosx = 1

⇒ x = 2π, 4π, 6π, 8π .......

And sin $\frac{5 x}{4} = 1$

⇒ x = $\frac{2 π}{5}, \frac{10 π}{5}, \frac{18 π}{5}$ ......

Common solution to above APs will be the AP having

First term = 2π

Common difference = LCM of 2π and $\frac{8 π}{5} = \frac{40 π}{5} = 8 π$

∴ General solution will be general term of this AP i.e. 2π + (8π)n, n ∈ I

⇒ x = 2(4n + 1)π, n ∈ I

5. Solve the equation (sinx + cosx)^1+sin2x= 2, when $0 \leq x \leq π$

Solution

We know, $- a^{2} + b^{2} \leq a sin θ + b cos θ \leq a^{2} + b^{2}$ and –1 ≤ sinθ ≤ 1.

∴ (sinx + cosx) admits the maximum value as $2$ and (1 + sin 2x) admits the maximum value as 2.

Also $(2)^{2} = 2$

∴ the equation could hold only when, sinx + cosx = $2$ and 1 + sin 2x = 2

Now, sinx + cos x = $2$

⇒ $cos (x - \frac{π}{4}) = 1$

⇒ x = 2nπ + $\frac{π}{4}$ , n ∈ I ...... (i)

and 1 + sin 2x = 2

⇒ sin2x = 1 = sin $\frac{π}{2}$

⇒ 2x = mπ + (–1)^m $\frac{π}{2}$ , m ∈ I

⇒ $x = \frac{mπ}{2} + (- 1)^{m} \frac{π}{4}$ ...... (ii)

The value of x in [0, π] satisfying equations (i) and (ii) is x = $\frac{π}{4}$

(when n = 0 & m = 0)

Note :- sin x + cos x = $- 2$ and 1 + sin 2x = 2 also satisfies but as x > 0, this solution is not in domain.

6. Find the set of values of x for which $\frac{t a n 3 x - t a n 2 x}{1 + t a n 3 x \cdot t a n 2 x} = 1.$

Solution

We have, $\frac{t a n 3 x - t a n 2 x}{1 + t a n 3 x \cdot t a n 2 x} = 1.$

⇒ tan (3x – 2x) = 1

⇒ tan x = 1

⇒ tan x = tan $\frac{π}{4}$

⇒ x = nπ + $\frac{π}{4}$ , n ∈ I {using tanθ = tanα ⇔ θ = nπ + α}

But for this value of x, tan 2x is not defined. Hence the solution set for x is φ.

6.0Also Read

Standard Units of Measurement	Angles	Introduction to Numbers
Ascending Order	Horizontal Line	Roman Numerals
Area of Irregular Shapes	Laws of Exponents	Rounding Numbers

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Strategies For Solving Trigonometric Equations

1.0Solutions Of Trigonometric Equations

2.0General Solutions of Some Trigonometric Equations

3.0Techniques for Simplifying and Solving Trigonometric Equations

Solving Trigonometric Equations by Factorisation

Solving Trigonometric Equations by Introducing an Auxiliary Argument

4.0Solved Examples

Solving Trigonometric Equation by Reducing it to a Quadratic Equation

Solving Trigonometric Equations by Introducing an Auxiliary Argument

5.0Techniques Of Trigonometric Equations Solution

6.0Also Read

Table of Contents

Strategies For Solving Trigonometric Equations

1.0Solutions Of Trigonometric Equations

2.0General Solutions of Some Trigonometric Equations

3.0Techniques for Simplifying and Solving Trigonometric Equations

Solving Trigonometric Equations by Factorisation

Solving Trigonometric Equations by Introducing an Auxiliary Argument

4.0Solved Examples

Solving Trigonometric Equation by Reducing it to a Quadratic Equation

Solving Trigonometric Equations by Introducing an Auxiliary Argument

5.0Techniques Of Trigonometric Equations Solution

6.0Also Read

Table of Contents