precalc hw, pls help

Question

1. Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree.

2cos^2θ−cosθ=0

2. Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree.

8sinxcosx=−5cosx

3. Assume that θ is a positive acute angle.

Given: cosθ= 12/37

Find: cos2θ

Answer 1

let x = cos(theta)

2x^2 - x = 0

x(2x-1) =0

x=0, x=1/2

cos(theta) = 0 or 1/2

theta =90 or 270 degrees, or pi/2 or 3pi/2

also 60 or 300 degrees or pi/3 or 5pi/3

8sin(theta)cos(theta) =-5cos(theta)

5cos(theta) + 8sin(theta)cos(theta) =0

cos(theta)[5 + 8sin(theta)] = 0

cos(theta) = 0 or sin(theta)=-5/8

theta = 90 or 270 degrees or 218.7 or 321.3 degrees

cos(theta) = 12/37

cos(2theta) = 2cos^2(theta) - 1 = 2(12/37)^2 -1 = 288/1369-1=-1081/1369

cos(2theta) = -1081/1369= about -0.7896

theta = about 71.075 degrees

2theta = 142.150

cos(142.15) = -0.7896