Solve the trigonometric equations over the interval [0,2pi)

Question

A. 3sec²xtanx=4tanx

B. 2-cosx=cos2x

Answer 1

A.) 3sec²tanx = 4tanx; tanx(3sec²x - 4) = 0; tanx = 0; x = 0 or x = π

3sec²x - 4 = 0; secx = 2/√3, x = π/6 or x = 11π/6

secx = - 2/√3, x = 5π/6 or x = 7π/6

B.) 2- cosx = 2cos²x -1; 2cos²x + cosx - 3 = 0

cosx = 1, x = 0

cosx = - 3/2 no solutions