Solve cos x = sin 2x, within the interval [0,2pi)?

1. Since cos 2x = cos^2 x - sin^ x, cos 4u = cos^2 2u - sin^2 2u, for all values of u in [0., 2π).

2. Again, an identity.

3. cos x = sin 2x

cos x = 2 sin x cos x

0 = 2 sin x cos x - cos x

0 = cos x(2 sin x - 1)

cos x = 0 or 2 sin x = 1

x = π/2, 3π/2......sin x = 1/2

............................x = π/6, 5π/6

4. cos^2 (x/2) = cos^2 x

(1 + cos x)/2 = cos^2 x

1 + cos x = 2 cos^2 x

0 = 2 cos^2 x - cos x - 1

0 = (2 cos x + 1)(cos x - 1)

2 cos x + 1 = 0 or cos x = 1

2 cos x = -1................x = 0

cos x = -1/2

x = 2π/3, 4π/3

3/

solve these

cosx = 0 and

sinx = 1/2