Question

4cos^2 theta+7cos theta=2

Answer 1

4cos^2 (theta) + 7 cos (theta) = 2
4cos^2 (theta) + 7 cos (theta) - 2 = 0
4cos^2 (theta) + 8 cos(theta) - cos(theta) - 2 = 0
(4cos^2 (theta) + 8cos(theta)) - (cos(theta) + 2) = 0
(4cos(theta))(cos(theta) + 2) - (1)(cos(theta) + 2) = 0
(4cos (theta) - 1)(cos (theta) + 2) = 0
cos(theta) = 1/4 or cos(theta) = -2 (rejected, since for all real values of theta, cos(theta) >= 1)

Thus, we have cos(theta) = 1/4.
Hence, the solution is theta = 2 n pi +/- cos^-1 (1/4), where n is any integer.