Originally Posted by

**pflo** Good start. There are probably simplier ways, but I'm a fan of being systematic. I always want to be dealing with the same angle in all my trig functions. I'd use the double-angle identities for both sine and cosine, then simplify and factor so I could use the zero product property to solve:

$\displaystyle 2\sin{x}\cos{x}+\cos{x}(1-\sin{x}^2)-\sin{x}(2\sin{x}\cos{x})=0$

$\displaystyle \cos{x}(2\sin{x}+1-\sin{x}^2-2\sin{x}^2)=0$

$\displaystyle \cos{x}(2\sin{x}+1-3\sin{x}^2)=0$

Applying the zero product property...

$\displaystyle \cos{x}=0$

$\displaystyle x=\arccos{x}$

$\displaystyle x=\pm\frac{\pi}{2}+2{\pi}k$

and...

$\displaystyle 3\sin{x}^2-s\sin{x}-1=0$

$\displaystyle 3\sin{x}+1=0$ and $\displaystyle \sin{x}-1=0$

Solve these (don't forget that you'll get multiple solutions like above with the cosine - then limit them to your interval).

Follow the same procedure for this problem. Get all the angles the same in your trig functions, then use the zero product property to solve. With the cosine double-angle identity, use $\displaystyle \cos{2t}=1-2\sin{t}^2$ instead of the other versions so that you'll have something to factor out (the $\displaystyle \sin{x}$) once you get it equal to zero.