# Solve the equation sin^2x=-2+5cosx

### 2 Answers | Add Yours

The equation `sin^2x=-2+5cosx` has to be solved.

`sin^2x=-2+5cosx`

Use the relation `sin^2x = 1 - cos^2x`

=> `1 - cos^2x = -2 + 5*cos x`

=> `cos^2x + 5*cos x - 3 = 0`

Let cos x = y

=> `y^2 + 5y - 3 = 0`

y1 = `(-5 + sqrt(25 + 12))/2 = -5/2 + sqrt 37/2`

y2 = `-5/2 - sqrt 37/2`

As y = cos x it has to lie in the set [-1, 1]

cos x = `-5/2 + sqrt 37/2`

x = `cos^-1((sqrt 37 - 5)/2)`

**The solution of the equation is **`x = cos^-1((sqrt 37 - 5)/2)`

starting frrom the equation:

`sin^2x=-2 + 5cos x`

add `cos^2x-1` both sides:

`sin^2+cos^2x -1=-2+5cosx +cos^2x -1`

so:

`cos^2x +5cosx-3= 0`

now we can rewrite equation:

`cos^2x +2 5/2 cosx +25/4 -37/4 = 0`

`(cosx +5/2)^2 = 37/4`

`cosx + 5/2=` `+-` `sqrt(37)/2`

so:

`cosx=-5/2` `+-` `sqrt(37)/2`

that is ; `cosx= 0,541381` `cosx= -5,541381`

of course the second solution isn't accetpted so:

`x = 57* 13' 20''`