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

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### 1 Answer

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`

Using quadratic formula, yields:

`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` .**

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