Give all answers in the interval 0 2 pi 2 sin theta Squarer

Give all answers in the interval (0, 2 pi) 2 sin theta = -Squareroot 3 2 sin (x) - 1 = 0 2 sin (x) - 1 = 0 cos theta + 3 cos theta - 4 = 0

1) 2sin = -sqrt(3)

sin = - sqrt(3)/2 = -sin(/3)

=> = ( + /3) = 4/3 or 2 - /3 = 5/3

2) 2sin²x - 1 = 0

sin²x = 1/2

sinx = +1/sqrt(2) or -1/sqrt(2)

When sinx = +1/sqrt(2) = sin /4

=> x = /4 or - /4 = 3/4

When sinx = -1/sqrt(2) = -sin/4

=> x = +/4 = 5/4 or 2 - /4 = 7/4

Hence x = /4, 3/4, 5/4, 7/4

3) same as second

4) cos² +3cos - 4 = 0

Solving the quadratic eqn we get

cos = -3 +/- sqrt(9 + 16)/2

=> cos = 1 or -4

but cos cannot be -4

hence cos = 1 = cos0 or cos2

=> = 0, 2