Mathematics Trigonometric Equations

### Topics Covered

star Introduction
star Theorem 1 { sin x = sin y }
star Theorem 2 {cos x = cos y}
star Theorem 3 {tan x = tan y}

### Introduction

\color{green} ✍️ Equations involving trigonometric functions of a variable are called color{blue}(ul"trigonometric equations.")

\color{green} ✍️ In this Section, we shall find the solutions of such equations.

\color{green} ✍️ We have already learnt that the values of sin x and cos x repeat after an interval of 2π and the values of tan x repeat after an interval of π.

\color{green} ✍️ The solutions of a trigonometric equation for which 0 ≤ x < 2π are called color{blue}(ul"principal solutions.")

\color{green} ✍️ The expression involving integer n which gives all solutions of a trigonometric equation is called color{blue}(ul"the general solution.")

### Theorem 1 { sin x = sin y }

For any real numbers x and y,

color{blue}(sin x = sin y) implies color{blue}(x = nπ + (–1)^n y) , where n ∈ Z

color{green}(ul"Proof : ")

If sin x = sin y, then

sin x – sin y = 0 or 2cos \ \(x+y)/2 sin \ \ (x-y)/2 = 0

which gives cos\ \ (x+y)/2 = 0 or sin \ \(x-y)/2 = 0

Therefore (x+y)/2 = (2n+1) pi/2  or (x-y)/2= npi where n ∈ Z

i.e. x = (2n + 1) π – y  or x = 2nπ + y, where n∈Z

Hence x = (2n + 1)π + (–1)^(2n + 1) y or x = 2nπ +(–1)^(2n) y, where n ∈ Z.

Combining these two results, we get

x = nπ + (–1)^n y, where n ∈ Z.
Q 3146001873

Find the principal solutions of the equation sin x = sqrt3/2

Solution:

We know that, sin( pi/3 )=(sqrt3)/2 and sin ((2pi)/3) = sin (pi - pi/3) = sin pi/3 = sqrt3/2

Therefore, principal solutions are x = pi/3 and (2pi)/3
Q 3116112070

Find the solution of sin x = – sqrt3/2

Solution:

We have sin x = – sqrt3/2 = - sin (pi/3) = sin (pi+pi/3) = sin 4pi/3

Hence sin x = sin ((4pi)/3) which gives

x = npi + (-1)^n (4pi)/3 where ∈ Z.
Q 3176212176

Solve 2 cos ^ 2 x + 3 sin x = 0

Solution:

The equation can be written as

2(1− sin^2 x) + 3sin x = 0

or 2 sin^2 x − 3sin x − 2 = 0

or (2sinx +1) (sinx − 2) =0

Hence sin x =- 1/2 or sin x = 2

But sin x = 2 is not possible

Therefore sin x = - 1/2 = sin ((7pi)/6)

Hence, the solution is given by

x = npi+(-1)^n (7pi)/6 where n ∈ Z.

### Theorem 2 {cos x = cos y}

For any real numbers x and y,

color{blue}(cos x = cos y), implies color{blue}(x = 2nπ ± y), where n ∈ Z

color{green}(ul"Proof :")

If cos x = cos y, then

cos x – cos y = 0 i.e.,

-2 sin \ \ (x+y)/2 sin \ \( x-y)/2 = 0

Thus sin \ \ (x+y)/2 = 0  or sin \ \(x-y)/2 = 0

Therefore x+y/2 = npi  or  x-y/2 = npi where n ∈ Z

i.e. x = 2nπ – y or x = 2nπ + y, where n ∈ Z

Hence x = 2nπ ± y, where n ∈ Z
Q 3146112073

Solve cos x = 1/2

Solution:

We have, cos x = 1/2 = cos pi/3

Therefore x = 2npi± pi/3 where n ∈ Z.
Q 3126212171

Solve sin 2x – sin4 x + sin 6x = 0.

Solution:

The equation can be written as
sin 6x + sin 2x − sin 4x = 0
or 2 sin 4x cos2x − sin 4x = 0
i.e. sin 4x(2 cos2x − 1) = 0

Therefore sin 4x = 0 or cos2x = 1/2

i.e. sin4x 0 or cos2x = cos pi/3

Hence 4x = npi or 2x = 2npi ± pi/3 where n∈Z

i.e. x = (npi)/4 or x = npi ± pi/6 where n∈Z.

### Theorem 3 {tan x = tan y}

Prove that if x and y are not odd multiple of pi/2 than

color{blue}(tan x = tan y) implies color{blue}(x = nπ + y) where n ∈ Z

color{green}(ul"Proof :")

If tan x = tan y, then tan x – tan y = 0

or (sin x cos y - cos sin y )/(cos x cos y) = 0

which gives sin (x – y) = 0

Therefore x – y = nπ, i.e., x = nπ + y, where n ∈ Z
Q 3166101975

Find the principal solutions of the equation tan x = - 1/sqrt3

Solution:

We know that, tan (pi/6) = 1/sqrt3 Thus , tan (pi - pi/6) = - tan( pi/6) = - 1/sqrt3

and tan(2pi - pi/6) = - tan pi/6 =- 1sqrt3

Thus tan( (5pi)/6) = tan ((11pi)/6)=- 1/sqrt3

Therefore, principal solutions are (5pi)/6 and (11pi)/6

We will now find the general solutions of trigonometric equations. We have already seen that:

sin x =0 gives x = nπ, where n ∈ Z

cos x =0 gives x = (2n+1)pi/2 where n ∈ Z.

We shall now prove the following results:
Q 3176112076

Solve tan 2x = - cot (x + pi/3)

Solution:

We have, tan 2x =- cot (x +pi/3) = tan (pi/2 + x+ pi/3)

or tan 2x = tan (x+(5pi)/6)

Therefore 2x = nx + x + (5pi)/6 where n∈Z

or x =npi + (5pi)/6 where n∈Z