Question

Solve 2 cos x + sec x = 3 on the interval [0,2pi)

Answer 1

recall secx=1/cosx
2cosx +1/cosx =3
2cos^2x +1 = 3cosx
2cos^2x-3cosx +1=0
Now the trick is to take y=cosx
2y^2-3y+1=0
solve this quadratic, then come back to y=cosx and reverse the roots you found , but keep the answers on [0,2pi]