trig intervals prob?

Question

solve on the interval 0 < equal to theta < 2pi


D) cos(2theta) = 2 - 2 sin^2 theta

E) cos (2theta) + 5 cos theta + 3 = 0


please help and explain, thanks!!

Answer 1

D) cos(2theta) = 2 - 2 sin^2 theta
1-2sin^2theta = 2-2sin^2theta
1=2 thus no solution

E) cos (2theta) + 5 cos theta + 3 = 0
2cos^2theta -1 +5costheta +3 =0
2cos^2theta +5costheta +2 = 0
cos^2theta +2.5costheta +1 =0
cos theta = [-2.5 +/- sqrt(6.25-4)]/2
costheta = -1.25 +/- .75
cos theta = - .5
theta = arccos (-.5) = 120 degrees and 210 degrees
theta = 2pi/3 or 7pi/6