Solve for 0 ≤ θ < 2π: -sq 2sinθ +2sinθcosθ −2sinθ =−2sinθ

Question

Solve for 0 ≤ θ < 2π:

2−sin^2 θ +cos^2 θ =3cosθ

0=−cos2θ +3cosθ −2

−cos2θ −2=3cosθ

would appreciate all the help for all 4 asap. Thank You

Answer 1

I cannot help you with the first one because I do not know what "sq" mean.

The remaining three all follow the same pattern:

You replace all occurrences of sin with cos using the formula

sin²θ = 1 - cos²θ

That will give you a quadratic equation in terms of cos.

2 − sin²θ + cos²θ = 3cosθ

2 - (1 - cos²θ) + cos²θ = 3cosθ

2cos²θ - 3cosθ + 1 = 0 (think of it as 2x² - 3x + 1 = 0)

cosθ = (3 ± √(9 - 8))/4 = 1, 1/2

θ = 0, π/3, 5π/3

0 = −cos2θ + 3cosθ − 2

0 = -(2cos²θ - 1) + 3cosθ − 2

2cos²θ - 3cosθ + 1 = 0

which is the same equation as above

−cos2θ − 2 = 3cosθ

-(2cos²θ - 1) - 2 = 3cosθ

2cos²θ + 3cosθ + 1 = 0 (think of it as 2x² + 3x + 1 = 0)

cosθ = (-3 ± √(9 - 8))/4 = -1, -1/2

θ = π, 2π/3, 4π/3