Mathematics Trigonometric Equations

Please Wait... While Loading Full Video

Class 11 Chapter 3 - Trigonometric Functions

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`