Solve `cos 2x gt | sin x|, x in (- pi/2, pi)`

Explanation

Draw the graphs of `y=cos2xandy=|sinx|`.

Let `cos2x=sinx`
`implies2sin^(2)x+sinx-1=0`
`impliessinx=-1,(1)/(2)`
But `sinxne-1impliessinx=(1)/(2)` Clearly, from the graphs of `y=|sinx|andy=cos2x,x=pm(pi)/(6),(5pi)/(6)`.
For `cos2xgt|sinx|`, the graph of `y=cos2x` must lie abvoe the graph of `y=|sinx|`.
From the graph the solution set is `x in(-(pi)/(6),(pi)/(6))uu((5pi)/(6),pi)`.