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. 

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) $\frac{\pi }{2}$ , n ∈ I

(c) If tan θ = 0, then θ = nπ, n ∈ I

(d) If sin θ = sin α, then θ = nπ + (–1)ⁿα ; where $\alpha \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ , 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 $\alpha \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

(g) If sin θ =1, then θ = 2nπ + $\frac{\pi }{2}$ = (4n + 1) $\frac{\pi }{2}$ ; 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 $\sin \frac{n\pi }{2}=(-1{)}^{\frac{n-1}{2}},\cos \frac{n\pi }{2}=0,$ $\sin \left(\frac{n\pi }{2}+\theta \right)=(-1{)}^{\frac{n-1}{2}}\cos \theta ,$ $\cos \left(\frac{n\pi }{2}+\theta \right)=(-1{)}^{\frac{n+1}{2}}\sin \theta$

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 = $\frac{1}{2}=sin\frac{\pi }{6}$

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

Solving Trigonometric Equations by Introducing an Auxiliary Argument

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

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

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

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

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

4.0Solved Examples

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

Solution

We have $\sqrt{3}\sin x+\cos x=\sqrt{3}$

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

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

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

$\Rightarrow \cos \left(x-\frac{\pi }{3}\right)=\cos \frac{\pi }{6}$

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

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

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

∴ $\cos \theta =\frac{1}{2}$

$\cos \theta =\cos \frac{\pi }{3}⟹\theta =2n\pi \pm \frac{\pi }{3},n\in \mathbb{I}$

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

$\theta =\frac{\pi }{3},\frac{5\pi }{3}$

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 =  $\frac{3\pi }{2}$ for which tanx + secx = 2 cosx  is not defined.

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

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

Solution

As  $\frac{n\pi }{2}$ ⇒ 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 $x\ne \frac{n\pi }{2}$

∴ No solution is possible.

5. Solve the equation sin⁴x + cos⁴ $\frac{7}{2}sinx.cosx.$

Solution

sin⁴x + cos⁴x = $\frac{7}{2}sinx.cosx$

⇒ (sin²x + cos²x)² – 2sin²x cos² $\frac{7}{2}sinx.cosx$

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

⇒ (2sin2x –1)(sin2x + 4) = 0

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

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

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

Solving Trigonometric Equations by Introducing an Auxiliary Argument

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

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

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

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

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

Solution

Here, sec x + tanx = $\sqrt{3}$

⇒ 1 + sinx =$\sqrt{3}$ cosx

or $\sqrt{3}$ cosx – sinx = 1

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

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

$\Rightarrow \cos \left(x+\frac{\pi }{6}\right)=\frac{1}{2}$

As 0 ≤ x ≤ 3π

$\frac{\pi }{6}\le x+\frac{\pi }{6}\le 3\pi +\frac{\pi }{6}$

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

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

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

∴ Total number of solutions are 2.

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

Solution

We have $\sqrt{3}\sin x+\cos x=\sqrt{3}$

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

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

$\Rightarrow \cos \left(x-\frac{\pi }{3}\right)=\cos \frac{\pi }{6}$

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

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

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

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 $\frac{\pi }{2}$ or sinx = $\frac{1}{2}$ = sin $\frac{\pi }{6}$

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

⇒ x = $\frac{2n\pi }{3}\pm \frac{\pi }{6}$

or x = mπ + (–1)^m $\frac{\pi }{6}$ ; (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, $cosA+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

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

⇒ 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 = $-\frac{1}{2}$

⇒ $\cos 2x=\cos \frac{\pi }{2}\text{or}\cos x=\cos \left(\pi -\frac{\pi }{3}\right)$

⇒ $\cos 2x=\cos \frac{\pi }{2}\text{or}\cos x=\cos \left(\frac{2\pi }{3}\right)$

⇒ $2x=(2n+1)\frac{\pi }{2}\text{or}x=2m\pi \pm \frac{2\pi }{3}$

⇒ $x=(2n+1)\frac{\pi }{4}\text{or}x=2m\pi \pm \frac{2\pi }{3}$

∴ the general solution is

x = (2n + 1) $\frac{\pi }{4}$ or 2mπ ± $\frac{2\pi }{3}$ , 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 = $\frac{n\pi }{2}$ , n ∈ I or 

2 cos 2x – 1= 0

⇒ cos2x = $\frac{1}{2}$

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

⇒ 2x = 2mπ ± $\frac{\pi }{3}$ ,  m ∈ I

⇒ x = mπ ± $\frac{\pi }{6}$ , m ∈ I

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

Solution

We have cos3x tan5x = sin7x 

⇒  $\cos 3x\left(\frac{\sin 5x}{\cos 5x}\right)=\sin 7x$

⇒ 2cos3x sin5x = 2cos 5x sin7x

⇒ sin8x + sin2x = sin12x + sin2x

⇒ sin12x – sin8x = 0

⇒ 2sin2x cos10x = 0

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

If cos 10x = 0, then

$10x=\frac{\pi }{2},\frac{3\pi }{2}$

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

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) $(\frac{\pi }{2})$ ,  n₁ ∈ I

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

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

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

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

Solution

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

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

∴ sin $(\frac{5x}{4})+cosx=2$

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

Now consider

cosx = 1

⇒ x = 2π, 4π, 6π, 8π .......

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

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

Common solution to above APs will be the AP having 

First term = 2π

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

∴ 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 $0\le x\le \pi$

Solution

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

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

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

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

Now, sinx + cos x = $\sqrt{2}$

⇒  $\cos \left(x-\frac{\pi }{4}\right)=1$

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

and  1 + sin 2x = 2

⇒ sin2x = 1 = sin $\frac{\pi }{2}$

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

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

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

(when n = 0 & m = 0)     

Note :- sin x + cos x = $-\sqrt{2}$ 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 $\frac{\tan 3x-\tan 2x}{1+\tan 3x\cdot \tan 2x}=1.$

Solution

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

⇒ tan (3x – 2x) = 1 

⇒  tan x = 1

⇒ tan x = tan $\frac{\pi }{4}$

⇒ x = nπ +$\frac{\pi }{4}$ , 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