Question

how do you solve cos2x = sinx when the directions tell you to find all solutions between the interval (0,2pi)?

Answer 1

By rewriting the above equation:
\[\cos(2x) = \cos(\frac{ \pi }{ 2 } - x)\]
and then solving:
\[2x = \pm(\pi/2 -x) +2*\pi*k (k = 0, \pm1, \pm2 ...)\]

Answer 2

You thus get 2 expressions:
\[2x = \pi/2 - x +2*\pi*k \]
and
\[2x = -\pi/2 + x + 2*\pi*k\]
which you need to solve for x and then substitute k =0, 1 -1 etc until you get results higher then 2pi or lower than 0