sec(4theta + 8 degrees) cos(6 theta + 2 degrees) = 1 find the least possible value of theta

Question

sec is 1/cos, so when you multiply them together do they cancel out and leave a quadratic equation for theta?

(4theta + 8) x (6theta +2) = 1 ?

Answer 1

sec (4θ + 8) cos (6θ + 2) = 1

cos (6θ + 2) = cos (4θ + 8)      multiply both sides by sec (4θ + 8)

6θ + 2 = 4θ + 8                      functions = if angles =

2θ + 2 = 8

2θ = 6

θ = 3

 

Radians or degrees?