What are trigonometric equations?

Trigonometric equations are equations involving trigonometric functions like sine, cosine, tangent, etc., where the goal is to find the values of the variables that satisfy the equation.

How to solve a trigonometric equation?

Trigonometric equations can be solved using various techniques, including algebraic manipulation, trigonometric identities, factoring, substitution, and the use of inverse trigonometric functions.

What are principal solutions in trigonometric equations?

Principal solutions are solutions within a specified interval, typically [0, 2π) or [0°, 360°) for angles measured in radians and degrees, respectively. These solutions represent the primary or fundamental angles satisfying the equation.

What are general solutions in trigonometric equations?

General solutions include all possible solutions to a trigonometric equation, accounting for the periodic nature of trigonometric functions. They are expressed as a principal solution plus integer multiples of the period of the trigonometric function.

How do I know if a solution to a trigonometric equation is valid?

To verify the validity of a solution, substitute it back into the original equation and ensure it satisfies the equation within the given domain or interval.

What are some common trigonometric identities used in solving equations?

Common trigonometric identities include Pythagorean identities, reciprocal identities, quotient identities, cofunction identities, sum and difference identities, and double-angle identities.

Can trigonometric equations have multiple solutions?

Yes, trigonometric equations can have multiple solutions within a given interval due to the periodic nature of trigonometric functions. It's important to consider all possible solutions when solving these equations.

What is the role of inverse trigonometric functions in solving equations?

Inverse trigonometric functions (e.g., arcsin, arccos, arctan) are used to find angles that satisfy trigonometric equations. Inverse trigonometric functions "undo" the trigonometric functions, allowing us to solve for the angle values.

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Trigonometric Equations

Trigonometric Equations involve expressions with trigonometric functions (like sine, cosine, tangent) set equal to a constant or another function. These equations are solved by using trigonometric ratios and identities, properties, and techniques such as factoring, substitution, or graphing. Solving trigonometric equations often involves finding solutions within a specific interval or generalizing solutions using periodicity properties of trigonometric functions.

1.0Trigonometric Equations

Trigonometric Equation Definition

In trigonometry, a trigonometric equation refers to an equation containing one or more trigonometric functions of unknown angles.

e.g. (i) $sin x = \frac{1}{2}$

(ii) cos²x – 4 sin x = 1

2.0Solution of Trigonometric Equation

A solution of a Trigonometric Equation refers to a value of the unknown angle that satisfies the given equation. It is the value that makes the equation true when substituted into it.

There are three types of Solution-

Principal Solution
Particular Solution
General Solution

These are described below-

Principal Solution:

The principal solution of a trigonometric equation is the solution that lies within a specific interval, typically [0, 2π) for angles measured in radians or [0°, 360°) for angles measured in degrees. This solution represents the primary or fundamental angle that satisfies the equation.

Particular Solution:

A particular solution of a trigonometric equation refers to a specific value or set of values for the variable(s) that satisfy the equation. It is a concrete solution within a given domain or interval that meets the conditions set forth by the equation.

General Solution:

The general solution of a trigonometric equation encompasses all possible solutions across multiple periods of the trigonometric functions involved. It typically includes the principal solution plus any integer multiples of the period of the trigonometric function. This yields an infinite set of solutions.

Also Check: Inverse Trigonometric Functions

Example 1: Find the principal solution for the equation $sin x = \frac{1}{2}$

Solution: $sin x = \frac{1}{2}$

$sin x = sin \frac{π}{6}$ or sin x = sin $\frac{5 π}{6}$

Hence $x = \frac{π}{6} or x = \frac{5 π}{6}$

3.0General Solution of Trigonometric Equation

Trigonometric Equation	General Solution
sin θ = 0	θ = nπ
cos θ = 0	θ = nπ + π/2
tan θ=0	θ = nπ
sin θ = 1	$θ = (2 nπ + \frac{π}{2}) = (4 n + 1) \frac{π}{2}$
cos θ = 1	θ = 2nπ
sin θ = sin α	θ = nπ + (–1)ⁿα, where α ∈ $[\frac{- π}{2}, \frac{- π}{2}]$
cos θ = cos α	θ = 2nπ ± α, where α ∈ (0, π]
tan θ = tan α	θ = nπ + α, where α ∈ $(\frac{- π}{2}, \frac{π}{2})$
sin 2θ = sin 2α	θ = nπ ± α
cos 2θ = cos 2α	θ = nπ ± α
tan 2θ = tan 2α	θ = nπ ± α

Question 1: Solve the trigonometric equation 3sin(x) – 2 = 0 for x in the interval [0,2π), and then provide its general solution.

Solution: To solve sin(x) – 2 = 0 in the interval [0,2π):

3sin(x) – 2 = 0

$x = sin^{- 1} (\frac{2}{3})$

Question 2: Solve: sec2θ = $- \frac{2}{3}$

Solution: ∵ sec 2θ= $- \frac{2}{3}$

⇒ cos2θ = $- \frac{3}{2}$ ⇒ cos2θ = cos $\frac{5 π}{6}$

⇒ 2θ = 2nπ ± $\frac{5 π}{6}$ , n ∈ Z ⇒ θ = nπ ± $\frac{5 π}{12}$ , n ∈ Z

Question 3: Solve 2 sin x $- 3$ = 0

Solution: 2 sin x = $3$

∵ sin x = $\frac{3}{2}$ ⇒ sin θ = sin $\frac{π}{3}$

∴ $θ = nπ + (- 1)^{n} \frac{π}{3}$ , n $\in$ Z

Question 4: Solve tan θ = $\frac{1}{3}$

Solution: ∵ tan θ = $\frac{1}{3}$

$tan θ = tan \frac{π}{6} \Rightarrow θ = n π + \frac{π}{6}, n \in Z$

4.0Principal Solution of Trigonometric Equation

The Principal Solution of a trigonometric equation is the solution within a specified interval, typically [0, 2π) for angles measured in radians or [0°, 360°) for angles measured in degrees. This solution represents the primary or fundamental angle satisfying the equation.

Example 1: Find the principal solutions of the following equation sin2θ = $- \frac{1}{2}$

Solution: $sin 2 θ = - \frac{1}{2}$

$Since, θ \in [2, 2 π), 2 θ \in [0, 4 π)$

$sin 2 θ = - \frac{1}{2}$

and $- \frac{1}{2} = - sin \frac{π}{6} = s in (π + \frac{π}{6}) = sin (2 π - \frac{π}{6}) = s in (3 π + \frac{π}{6}) = s in (4 π - \frac{π}{6}) ..$

[∵ sin (π + θ) = sin (2π – θ) = sin (3π + θ) = sin (4π – θ) = – sin θ]

$∴ sin 2 θ = sin \frac{7 π}{6} = sin \frac{11 π}{6} = sin \frac{19 π}{6} = sin \frac{23 π}{6}$

$∴ 2 θ = \frac{7 π}{6} or 2 θ = \frac{11 π}{6} or 2 θ = \frac{19 π}{6} or 2 θ = \frac{23 π}{6}$

$∴ θ = \frac{7 π}{12} or θ = \frac{11 π}{12} or θ = \frac{19 π}{12} or θ = \frac{23 π}{12}$

Hence, the required principal solutions are ${\frac{7 π}{12}, \frac{11 π}{12}, \frac{19 π}{12}, \frac{23 π}{12}}$

Example 2: Find the principal solutions of the following equations: $s in x = \frac{1}{2}$

$(A) \frac{π}{6} (B) \frac{π}{6}, \frac{5 π}{6} (C) \frac{π}{6}, - \frac{π}{6} (D)$ None of these

Solution:

The Correct Answer is (B)

The given equation is $x = \frac{1}{2}$

We know that, $s in \frac{π}{6} = \frac{1}{2} an d (π - \frac{π}{6}) = \frac{1}{2}$

$∴ sin \frac{π}{6} = \frac{1}{2} and sin \frac{5 π}{6} = \frac{1}{2} .$

Hence, the principal solution is, $x = \frac{π}{6} an d x = \frac{5 π}{6} .$

Also Check: Trigonometry previous year questions with solutions

5.0How to Find the Roots of a Trigonometric Equations

Example 1: $6 - 10 cos x = 3 sin^{2} x$

Solution: $6 - 10 cos x = 3 - 3 cos^{2} x \Rightarrow 3 cos^{2} x - 10 cos x + 3 = 0$

⇒ $(3 cos x - 1) (cos x - 3) = 0 \Rightarrow cos x = 3 or cos x = 3$

Since cos x = 3 is not possible as – 1 ≤ cos x ≤ 1

∴ $cos x = \frac{1}{3} = cos (cos^{- 1} \frac{1}{3}) \Rightarrow x = 2 nπ \pm cos^{- 1} (\frac{1}{3}), n \in Z$

Example 2: Solve $sin^{2} θ - cos θ = \frac{1}{4}$ for θ and write the values of in the interval 0 ≤ θ ≤ 2 π.

Solution: The given equation can be expressed as

$1 - cos^{2} θ - cos θ = \frac{1}{4}$

$\Rightarrow cos^{2} θ + cos θ - \frac{3}{4} = 0$

$\Rightarrow 4 cos^{2} θ + 4 cos θ - 3 = 0$

$\Rightarrow (2 cos θ - 1) (2 cos θ + 3) = 0$

$\Rightarrow cos θ = \frac{1}{2}, - \frac{3}{2}$

$cos θ = - 2 is not possible as - 1 \leq cos θ \leq 1$

$∴ cos θ = \frac{1}{2} \Rightarrow cos θ = cos \frac{π}{3} \Rightarrow θ = 2 nπ \pm \frac{π}{3}, n \in Z$

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

$\Rightarrow θ = \frac{π}{3}, \frac{5 π}{3}$

6.0Solving Linear Trigonometric Equations

Example 1: $(2 sin x - cos x) (1 + cos x) = sin^{2} x$

Solution: $∴ (2 sin x - cos x) (1 + cos x) - (1 - cos^{2} x) = 0$

$∴ (1 + cos x) (2 sin x - cos x - 1 + cos x) = 0$

$∴ (1 + cos x) (2 sin x - 1) = 0$

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

$\Rightarrow cos x = - 1 = cos π \Rightarrow x = 2 nπ \pm π = (2 n + 1) π, n \in Z$

or $sin x = \frac{1}{2} = sin \frac{π}{6} \Rightarrow x = kπ + (- 1)^{k} \frac{π}{6}, k \in Z$

7.0Trigonometric Equations Examples with Solutions

Example 1: cos 3x + sin 2x – sin 4x = 0

Solution: cos 3x – 2 sin x cos 3x=0

⇒ (cos3x) (1 – 2sinx) = 0

$\Rightarrow cos 3 x = 0 or sin x = \frac{1}{2}$

$\Rightarrow cos 3 x = 0 = cos \frac{π}{2} or sin x = \frac{1}{2} = sin \frac{π}{6}$

$\Rightarrow 3 x = 2 nπ \pm \frac{π}{2} or x = mπ + (- 1)^{m} \frac{π}{6}$

$\Rightarrow x = \frac{2 nπ}{3} \pm \frac{π}{6} or x = m π + (- 1)^{m} \frac{π}{6}; (n, m \in Z)$

Example 2: Solve cosθ + cos3θ + cos5θ + cos7θ=0

Solution: We have cosθ + cos7θ + cos3θ + cos5θ = 0

⇒ 2 cos4θ cos3θ + 2 cos4θ cosθ = 0 ⇒ cos4θ(cos3θ + cosθ) = 0

⇒ cos4θ(2cos2θcosθ) = 0

$\Rightarrow Either cos θ = 0 \Rightarrow θ = (2 n_{1} + 1) \frac{π}{2}, n_{1} \in Z$

$O r cos 2 θ = 0 \Rightarrow θ = (2 n_{2} + 1) \frac{π}{4}, n_{2} \in Z$

$O r or cos 4 θ = 0 \Rightarrow θ = (2 n_{3} + 1) \frac{π}{8}, n_{3} \in Z$

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Trigonometric Equations

Trigonometric Equations involve expressions with trigonometric functions (like sine, cosine, tangent) set equal to a constant or another function. These equations are solved by using trigonometric ratios and identities, properties, and techniques such as factoring, substitution, or graphing. Solving trigonometric equations often involves finding solutions within a specific interval or generalizing solutions using periodicity properties of trigonometric functions.

1.0Trigonometric Equations

Trigonometric Equation Definition

In trigonometry, a trigonometric equation refers to an equation containing one or more trigonometric functions of unknown angles.

e.g. (i) $sin x = \frac{1}{2}$

(ii) cos²x – 4 sin x = 1

2.0Solution of Trigonometric Equation

A solution of a Trigonometric Equation refers to a value of the unknown angle that satisfies the given equation. It is the value that makes the equation true when substituted into it.

There are three types of Solution-

Principal Solution
Particular Solution
General Solution

These are described below-

Principal Solution:

The principal solution of a trigonometric equation is the solution that lies within a specific interval, typically [0, 2π) for angles measured in radians or [0°, 360°) for angles measured in degrees. This solution represents the primary or fundamental angle that satisfies the equation.

Particular Solution:

A particular solution of a trigonometric equation refers to a specific value or set of values for the variable(s) that satisfy the equation. It is a concrete solution within a given domain or interval that meets the conditions set forth by the equation.

General Solution:

The general solution of a trigonometric equation encompasses all possible solutions across multiple periods of the trigonometric functions involved. It typically includes the principal solution plus any integer multiples of the period of the trigonometric function. This yields an infinite set of solutions.

Also Check: Inverse Trigonometric Functions

Example 1: Find the principal solution for the equation $sin x = \frac{1}{2}$

Solution: $sin x = \frac{1}{2}$

$sin x = sin \frac{π}{6}$ or sin x = sin $\frac{5 π}{6}$

Hence $x = \frac{π}{6} or x = \frac{5 π}{6}$

3.0General Solution of Trigonometric Equation

Trigonometric Equation	General Solution
sin θ = 0	θ = nπ
cos θ = 0	θ = nπ + π/2
tan θ=0	θ = nπ
sin θ = 1	$θ = (2 nπ + \frac{π}{2}) = (4 n + 1) \frac{π}{2}$
cos θ = 1	θ = 2nπ
sin θ = sin α	θ = nπ + (–1)ⁿα, where α ∈ $[\frac{- π}{2}, \frac{- π}{2}]$
cos θ = cos α	θ = 2nπ ± α, where α ∈ (0, π]
tan θ = tan α	θ = nπ + α, where α ∈ $(\frac{- π}{2}, \frac{π}{2})$
sin 2θ = sin 2α	θ = nπ ± α
cos 2θ = cos 2α	θ = nπ ± α
tan 2θ = tan 2α	θ = nπ ± α