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

Express the equation in one trigonometric function. To do so, replace `sin^2x` by `1-cos^2x` . This is base on the Pythagorean identity `sin^2x+cos^2x=1` .

`1-cos^2x-5cos x = -2`

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

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

Let cosx = z to express the equation in a form `ax^2+bx+c=0` .

`-z^2 - 5z + 3 = 0`

Then, use the quadratic formula to solve for z.

`z=(-b+-sqrt(b^2-4ac))/(2a) = [-(-5)+-sqrt((-5)^2-(4)(-1)(3))]/(2*(-1))`

`z= [5+-sqrt(25+12)]/(-2)= (5+-sqrt37)/(-2)`

`z_1 = (5+sqrt37)/(-2)=-5.54`

`z_2=(5-sqrt37)/(-2)=0.54`

Then, substitute the values of z to cos x=z.

`z_1=-5.54` , `cos x = -5.54` *(Invalid cosine function)*

`z_2=0.54` , `cos x = 0.54`

Note that we cannot consider the first value of z. The range of a cosine function varies from -1 to 1 only. So the values of z should be in the interval `-1lt=zlt=1` .

Then,

`cos x =0.54`

`x = cos^(-1) (0.54)`

Since we have a positve cosine, angle x is located at the first and fourth quadrant of the unit circle.

x = 57.32 and 302.68 degrees

**The general solution of the equation `sin^2x -5cosx =-2` are:**

**`x_1 = 57.32+ 360k ` degrees and**

**`x_2 = 302.68 + 360k` degrees **

**where k is an integer .**

The equation `sin^2x - 5*cos x = -2` has to be solved.

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

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

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

Let cos x = y

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

`y1 = (-5 + sqrt 37)/2` and `y2 = (-5 - sqrt 37)/2`

Eliminate `(-5 - sqrt 37)/2` as it is smaller than -1

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

`~~ 0.9987`

and `x ~~5.284`

**The solution of the equation is **`0.9987 + 2*n*pi` and `5.284 +2*n*pi`