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## Answers (3)

cos²x - sin²x - 2sinx = 0

(1-sin²x) - sin²x - 2sinx = 0

1-2sin²x - 2sinx = 0

2sin²x + 2sinx - 1 = 0

sinx = (1/4)[-2±√(4+8)] = (-1±√3)/2

discard the sinx = -(1+√3)/2 since |sinx| ≤ 1

x = arcsin[(√3 - 1)/2] ≈ 21.471° ± 360n, 158.529° ± 360n (n = 0, ±1, ±2, ±3, ... )

...

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