The equation `2*sin^2 x + cos x - 1 = 0` has to be solved for x.

This trigonometric equation has both sin x as well as cos x. It is possible to write `sin^2x = 1 - cos^2x` . This eliminates sin x and would allow the equation to be written as a quadratic in terms of cos x.

`2*sin^2 x + cos x - 1 = 0`

=> `2*(1 - cos^2x) + cos x - 1 = 0`

=> `2 - 2*cos^2x + cos x - 1 = 0`

=> `-2*cos^2x + cos x + 1 = 0`

=> `2*cos^2x - cos x - 1 = 0`

=> `2*cos^2x - 2*cos x + cos x - 1 = 0`

=> `2*cos x(cosx - 1) + 1(cos x - 1) = 0`

=> `(2*cos x + 1)(cos x - 1) = 0`

`2*cos x + 1 = 0 => cos x = -1/2`

`cos x - 1 = 0 => cos x = 1`

The cosine function has a periodicity of `2*pi` . This gives an infinite number of solutions of the equation.

`cos x = -1/2`

=> `x = 2*pi/3 + 2*n*pi` and `x = 4*pi/3 + 2*n*pi`

cos x = 1

=> `x = 2*n*pi`

The solutions of the equation `2*sin^2 x + cos x - 1 = 0` are `{2*pi/3 + 2*n*pi, 4*pi/3 + 2*n*pi, 2*n*pi}`

To solve the equation `2*sin^2 x + cos x - 1 = 0` first change `sin^2x` to `cos^2x ` so that the equation can be written in the form of a quadratic equation.

`2*sin^2 x + cos x - 1 = 0`

Use the formula `sin^2x = 1 - cos^2 x` , this gives:

`2*(1 - cos^2x) + cos x - 1 = 0`

`2 - 2*cos^2x + cos x - 1 = 0`

`-2*cos^2x + cos x + 1 = 0`

If `y = cos^2x` , the given equation is `-2y^2 + y + 1 = 0`

The solution of this quadratic equation is `(-1+-sqrt(1 + 8))/(-4)`

= `(-1+- 3)/(-4)`

= `-1/2` and 1

cos x = 1/2 gives x = 120 + n*360, x = 240 + n*360

cos x = 1 gives x = n*360

The solution of the equation is 120+n*360, 240+n*360 and n*360