trigonometry prob help?

Question

Solve. Give all positive values of the angle between 0º and 360º that will satisfy each. Give any approximate value to the nearest minute only.


6. cos 2x - sin² x/2 + ¾ = 0


7. 3 sin theta - 4 cos theta = 2

Answer 1

cos 2x - sin² x/2 + ¾ = 0
sin^2 x = (1-cos 2x)/2
cos 2x - sin² x/2 = cos2x -(1-cos2x)/2
= (cos2x-1)/2
=> (cos2x-1)/2 + 3/4 = 0
=> cos2x -1 = -3/2
cos2x = -1/2
2x = 120
x = 60

Answer 2

6. cos(2x) - sin²(x/2) + ¾ = 0

cos(2x) - sin²(x/2) + ¾ = 0
(2cos²x - 1) - (1 - cosx)/2 + ¾ = 0
2cos²x - 1 - 1/2 + cosx/2 + ¾ = 0
2cos²x + cosx/2 - ¾ = 0
8cos²x + 2cosx - 3 = 0
(4cosx + 3)(2cosx - 1) = 0
cosx = -3/4, 1/2
x = {arccos(-3/4)°, 60°,[360 - arccos(-3/4)]°, 300°}


7. 3sinθ - 4cosθ = 2

3sinθ - 4cosθ = 2
3sinθ = 4cosθ + 2
(3sinθ)² = (4cosθ + 2)²
9sin²θ = 16cos²θ + 16cosθ + 4
9 - 9cos²θ = 16cos²θ + 16cosθ + 4
0 = 25cos²θ + 16cosθ - 5
25cos²θ + 16cosθ - 5 = 0
cosθ = {-16 ± √(256 + 4*25*5)}/50 = {-16 ± √756}/50

However one invalid solution was introduced in the squaring and must be rejected. That leaves us with

cosθ = {-16 + √756}/50

cosθ ≈ 0.229909
θ ≈ 76.708281°,283.29172°