# HOW TO SOLVE TRIGONOMETRIC EQUATIONS

How to Solve Trigonometric Equations :

Solving trigonometrical equations is nothing but finding all replacements for the variable θ that make the equations true.

A solution of a trigonometrical equation is the value of the unknown angle that satisfies the equation. A trigonometrical equation may have infinite number of solutions.

Principal Solution :

The solution in which the absolute value of the angle is the least is called principal solution.

General Solution :

The expression involving integer ‘n’ which gives all solutions of a  trigonometrical equation is called the general solution.   ## How to Solve Trigonometric Equations - Examples

Example 1 :

Solve  2 cos²θ + 3 sin θ = 0

Solution :

2 cos²θ + 3 sin θ = 0

First we have to convert  the given function in one trigonometric ratio.

cos²θ = 1 - sin²θ

2 (1 - sin²θ) + 3 sin θ = 0

2 - 2 sin²θ + 3 sin θ = 0

Multiply the whole equation by (-)

2 sin²θ - 3 sin θ - 2 = 0

2 sin²θ - 4 sin θ + sin θ - 2 = 0

2 sin θ (sin θ - 2) + 1(sin θ - 2) = 0

(2 sin θ + 1) (sin θ - 2) = 0

 2 sin θ + 1 = 0Subtract 1 on both sides2 sin θ + 1 - 1 = 0 - 12 sin θ = -1Divide by 2 on both sidessin θ = -1/2 sin θ - 2 = 0sin θ =  2it is not possible

sin θ = - sin (π/6)

sin θ = sin (-π/6)  [satisfies the condition sin θ = sin α]

So, α = θ = -π/6

Other angles are

General angle for sin θ :

θ = nπ + (-1)ⁿ  α    [n ∈ z ]

θ = nπ + (-1)ⁿ (-π/6)

For example

if we put n = 1, we will get

θ = π + (-1)¹ (-π/6) ==> π -π/6 ==> 5π/6

Hence θ = -π/6, 5π/6, ..........

Example 2 :

Solve 2 tan θ − cot θ = − 1

Solution :

2 tan θ − cot θ = − 1

2 tan θ − (1/tan θ) = − 1

(2 tan²θ − 1)/tan θ = − 1

Multiply by tan θ on both sides

(2 tan²θ − 1) = − tan θ

2 tan²θ  + tan θ − 1 = 0

Let x = tan θ

2x² + x - 1 = 0

2x² + 2x - x - 1 = 0 ==> 2x(x + 1) - 1(x + 1) = 0

(2x - 1) (x + 1) = 0

 2 x - 1 = 02 tan θ - 1 = 0Add 1 on both sides2 tan θ = 1Divide by 2 on both sidestan θ = 1/2 x + 1 = 0Subtract 1 on both sidesx + 1 - 1 = 0 - 1x = -1tan θ = -1

tan θ = -1

tan θ = -tan (π/4)

tan θ = tan (-π/4)  [satisfies the condition tan θ = tan α]

So, α = θ = -π/4

Other angles are

General angle for tan θ :

θ = nπ + (-1)ⁿ  α    [n ∈ z ]

θ = nπ -(π/4)

tan θ = 1/2

θ = tan ¹(1/2)

θ = nπ + (-1)ⁿ  α    [n ∈ z ]

θ = nπ + tan ⁻¹(1/2) After having gone through the stuff given above, we hope that the students would have understood how to solve trigonometric equations.

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