# Solve equation 3(cos x)^2-sin 2x-(sin x)^2=0 if 0 degrees<x<90 degrees?

sciencesolve | Certified Educator

You should notice that the members of the equation do not have the same argument, hence, you may use the double angle identity for the middle term, such that:

`3cos^2 x - 2sin x*cos x - sin^2 x = 0`

Since all the members of the equation have the same degree, hence, the equation becomes homogeneous and you may divide the equation by `cos^2 x` , such that:

`3 - 2(sin x/cos x) - (sin^2 x)/(cos^2 x) = 0`

Using the trigonometric identity `tan x = sin x/cos x` yields:

`3 - 2tan x - tan^2 x = 0`

`-tan^2 x - 2tan x + 3 = 0 => tan^2 x + 2tan x - 3 = 0 `

You may come up with the following substitution, such that:

`tan x = y`

`y^2 + 2y - 3 = 0`

`y_(1,2) = (-2+-sqrt(4+12))/2 => y_(1,2) = (-2+-4)/2`

`y_1 = 1 ; y_2 = -3`

You need to solve the equations, such that:

`tan x = y_1 => tan x = 1 => x = pi/4 = 45^o in (0^o,90^o)`

`tan x = -3` invalid since the value of tangent needs to be positive in quadrant 1

Hence, evaluating the solution to the given homogeneous trigonometric equation, yields `x = 45^o` .