# `2cos^2x+sinx.cosx-sin^2x=0`

### 1 Answer | Add Yours

`2cos^2x + sinxcosx - sin^2x = 0`

To solve for x, consider the above expression as a quadratic equation. Then, factor.

`(2cosx - sinx) ( cosx + sinx) = 0`

Set each factor to zero and solve for x.

> `2cosx - sinx = 0`

Divide both sides by cos x.

`2 - (sinx)/(cosx) = 0`

`tanx = 2`

Note that tangent function is positive at quadrants I and III. So,

`x = 63.43` and `243.43` degrees

The other factor is:

> `cos x + sinx = 0`

Divide both sides by cos x.

`1 + sinx/cosx = 0`

`1 + tanx = 0`

`tan x = -1`

Tangent function is negative at quadrant II and IV. The values of angle x:

`x = 135` and `315` degrees

**Hence, the general solution of `2cos^2x + sinxcosx-sin^2x` are:**

**`x_1 = 63.43 + 360k`degrees**

**`x_2 = 135 + 360k` degrees**

**`x_3 = 243.43 + 360k` degrees and**

**`x_4 = 315 +360k` degrees**