Question

Question #a6bad

Answer 1

There is no real solution.

Explanation:

Let `S = sin^2x`. Use `cos 2x=1-2S`.

The given equation becomes

`S-(1-2S)^2+3/4`

`-1/4(16S^2-20S+1)=0` giving solutions

`S=sin^2x=(5+-4sqrt6)/4= -1.2 < -1 and 3.7 >1`.

As `|sin^2x|<=1`, there is no real solution.