#### notes

Equations involving trigonometric functions of a variable are called trigonometric equations. The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solutions. The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution.

How will solve a trigonometric equation `2"sin"^2 x+ 7"cos" x= 5`? We shall follow few steps:**1) Reduce variables-** Try to reduce the equation to one variable.

`2(1- cos^2 x)+ 7cos x= 5`**2) Simplify equation**

`2cos^2 x- 7cos x+ 3= 0`

`2cos^2 x- 6cos x- cos x+3= 0`

`2cos x (cos x-3)-1 (cos x- 3)= 0`

`(2cos x- 1) (cos x- 3)= 0`

`cos x= 1/2` or `cos x= 3`**3) Double check domain and range of varibles**

The trigonometric function cos lies between -1, 1

therefore, cos x= `1/2`

we know that cos 60°= `1/2`

thus, x= 60°**4) Find all possible solutions**

Principal soultions- 0 ≤ x ≤ 2π

60°, 300°

General solutions-

x= n(2π) ± y

x= 2nπ ± y

#### theorem

**Theorem 1:** For any real numbers x and y,

sin x = sin y implies x = nπ + (–1)n y, where n ∈ Z

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) π/2` or `(x-y)/2= nπ`, 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.

**Theorem 2:** For any real numbers x and y, cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

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= nπ` or `(x-y)/2= nπ`, where n ∈ Z

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

**Theorem 3:**Prove that if x and y are not odd mulitple of `π/2`, then

tan x = tan y implies x = nπ + y, where n ∈ Z

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

or `(sin x cos y- cos x 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.