9cosx + 4cos(2x) + 1 = 0
9cosx + 4(2cos^2 x - 1) + 1 = 0
8cos^2 x + 9cosx - 3 = 0
now just solve the quadratic equation for cosx
Solve rounding to 2 decimal places.
9cosx + 4cos(2x) + 1 = 0
over 0° ≤ x < 360°
Thanks to anyone that can help! I'm totally stumped.
1 answer