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 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

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-

1. Principal Solution
2. Particular Solution
3. General Solution

These are described below-

1. 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.

2. 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.

3. 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.

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

Solution: $\sin x=\frac{1}{2}$ 

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

Hence  $x=\frac{\pi }{6}orx=\frac{5\pi }{6}$

3.0General Solution of Trigonometric Equation 

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

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

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

Solution: ∵ sec 2θ= $-\frac{2}{\sqrt{3}}$

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

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

Question 3: Solve 2 sin x$-\sqrt{3}$ = 0

Solution: 2 sin x = $\sqrt{3}$

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

∴ $\theta =n\pi +(-1{)}^n\frac{\pi }{3}$, n $\in$ Z

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

Solution: ∵ tan θ = $\frac{1}{\sqrt{3}}$

$\tan \theta =\tan \frac{\pi }{6}\Rightarrow \theta =\mathrm{n}\pi +\frac{\pi }{6},\mathrm{n}\in \mathrm{Z}$

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: $\sin 2\theta =-\frac{1}{2}$

$\text{ Since, }\theta \in [2,2\pi ),2\theta \in [0,4\pi )$

$\sin 2\theta =-\frac{1}{2}=-\sin \frac{\pi }{6}=\sin \left(\pi +\frac{\pi }{6}\right)=\sin \left(2\pi -\frac{\pi }{6}\right)=\sin \left(3\pi +\frac{\pi }{6}\right)=\sin \left(4\pi -\frac{\pi }{6}\right)\dots \dots .[\because sin(\pi +\theta )=sin(2\pi –\theta )=sin(3\pi +\theta )=sin(4\pi –\theta )=–sin\theta ]$

$\therefore \sin 2\theta =\sin \frac{7\pi }{6}=\sin \frac{11\pi }{6}=\sin \frac{19\pi }{6}=\sin \frac{23\pi }{6}$

$\therefore 2\theta =\frac{7\pi }{6}\text{ or }2\theta =\frac{11\pi }{6}\text{ or }2\theta =\frac{19\pi }{6}\text{ or }2\theta =\frac{23\pi }{6}$

$\therefore \theta =\frac{7\pi }{12}\text{ or }\theta =\frac{11\pi }{12}\text{ or }\theta =\frac{19\pi }{12}\text{ or }\theta =\frac{23\pi }{12}$

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

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

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

Solution:

The Correct Answer is (B)

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

We know that, $sin\frac{\pi }{6}=\frac{1}{2}and\left(\pi -\frac{\pi }{6}\right)=\frac{1}{2}$

$\therefore \sin \frac{\pi }{6}=\frac{1}{2}\text{ and }\sin \frac{5\pi }{6}=\frac{1}{2}\text{. }$

Hence, the principal solution is, $x=\frac{\pi }{6}andx=\frac{5\pi }{6}.$

5.0How to Find the Roots of a Trigonometric Equations

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

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

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

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

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

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

Solution: The given equation can be expressed as

$1-{\cos }^2\theta -\cos \theta =\frac{1}{4}$

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

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

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

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

$\cos \theta =-2\text{ is not possible as }-1\le \cos \theta \le 1$

$\therefore \cos \theta =\frac{1}{2}\Rightarrow \cos \theta =\cos \frac{\pi }{3}\Rightarrow \theta =2n\pi \pm \frac{\pi }{3},\mathrm{n}\in \mathrm{Z}$

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

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

6.0Solving Linear Trigonometric Equations

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

Solution:  $\therefore (2\sin x-\cos x)(1+\cos x)-\left(1-{\cos }^{2}x\right)=0$

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

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

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

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

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

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

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

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

$\Rightarrow 3x=2n\pi \pm \frac{\pi }{2}\text{ or }x=m\pi +(-1{)}^m\frac{\pi }{6}$

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

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

$\Rightarrow \text{ Either }\cos \theta =0\Rightarrow \theta =\left(2n_1+1\right)\frac{\pi }{2},n_1\in \mathrm{Z}$

$Or\cos 2\theta =0\Rightarrow \theta =\left(2n_2+1\right)\frac{\pi }{4},n_2\in Z$

$Or\text{ or }\cos 4\theta =0\Rightarrow \theta =\left(2{\mathrm{n}}_3+1\right)\frac{\pi }{8},{\mathrm{n}}_3\in \mathrm{Z}$