how do you solve (cosx)^2-(sinx)^2-2sinx=0?

Using the fact that (sin x)^2 + (cos x)^2 = 1

Replace (cos x)^2 with (1 - (sin x)*2)

Which gets you 1 - 2(sin x)^2 - 2sin x = 0

Now realize that this just a 2nd degree equation. If you replace sin x with z you have

1 - 2z^2 - 2z =0

Use quadratic equation to solve for z. And since z = sin x, take inverse sine of z to get x

(cosx)^2-(sinx)^2-2sinx=0

1-(sinx)^2-(sinx)^2-2sinx=0

2(sinx)^2+2sinx -1 =0

sin x = (-1+sqrt(3))/2 or {sin x = (-1- sqrt(3))/2 not allowed}

then only

sin x = (-1+sqrt(3))/2

x = (-1)^k*arcsin((-1+sqrt(3))/2)+kpi where k any integer