Find all solutions in the interval 0° ≤ 𝜃 < 360°

Question

4 cos 2𝜃 - 2 sin𝜃= 3

Answer 1

4cos2x -2sinx = 3

4(cos^2(x) -sin^2(x)) -2sinx = 3

2(cos^2(x) +sin^2(x)) -sinx = 3/2 + 4sin^2(x)

1 -sinx = 3/2 + 4sin^2(x)2

4sin^2(x) + sinx -1/2 = 0

sin(x) = -1/8 + or - (1/8)sqr(1+8)

sin(x) =-1/8 + or - 3/8 = -1/2, 1/4

x = 11pi/6, 7pi/6, .08pi, or .92pi

x = sin^-1(.1124) = about 6.45 degrees

4cos2(6.45)- 2(sin6.45)

3.9 -.22 =