trig. equation?

Question

find the solutions of the equation in the interval [0,2pi)

sinx/2-cosx=0

Answer 1

sinx/2 = cosx
sinx = 2cosx
(sinx)^2 = 4(cosx)^2
(sinx)^2 = 4(1-sinx)^2
(sinx)^2 = 4 - (4sinx)^2
5(sinx)^2 = 4
(sinx)^2 = 0.8
sinx = 0.8^(1/2)
x = sin^(-1) (0.8^(1/2))
= 63.4 (3sf)

Answer 2

sinx/2-cosx=0
sqrt((1-cosx)/2) = cosx
1-cosx = 2cos^2 x
2cos^2x +cosx -1 = 0
(2cosx-1)(cosx+1) = 0
cos x = 1/2 --> x = 60 or 300 degrees
cosx = -1 --> x = 180 degrees is rejected
So x = pi/6, 5pi/2