Question

. Find, to the nearest degree, all values of θ in the interval 0°  θ  180° that satisfy the equation 8cos 2 θ – 2 cos θ – 1 = 0.

Answer 1

y = 8cos 2x - 2c0s x - 1 = 0. Replace cos 2x by (2cos^2 x - 1)
16 cos^2 x -8 - 2cos x - 1 = 0. Call cos x = x, solve the quadratic equation:
16x^2 - 2x - 9 = 0
D = b^2 - 4ac = 4 + 576 = 580 --> VD = 24.08
x1 = 4/32 + 24.08/32 = 0.878
x2 = 4/32 - 24.08/32 = -0.628
a) cos x = x = 0.878
Calculator gives cos x1 = cos 28.60 deg -> x1 = 28.60 deg
Unit circle gives another cos x = cos (360 - 28.60) = 331.40 -> x2 = 331.40
cos
b) cos x = -0,628 = cos (180 + 51.10) -> x3 = 231.10 deg
Unit circle gives another cos x = cos (180 - 51.10) = cos 128.9-> x4 = 128.9
Finally,inside the interval (0, 180), there are 2 answers
x1 = 28.60 deg, and x3 = 128.90 deg.
Inside the interval (0, 360) there are 4 answers.