Hi there,

I'm having a little trouble solving the following equation for $\displaystyle -\pi <x\leq 2 \pi$:

$\displaystyle sin^2(x)cos^3(x)=cos(x)$

I've rearranged it into the form:

$\displaystyle cos(x)\left [sin^2(x)cos^2(x)-1\right ]=0$

So where $\displaystyle cos(x)=0,\ \ \ x= \pm\frac{\pi}{2}$ and $\displaystyle \frac{3\pi}{2}$

But I'm just not sure how to find $\displaystyle x$ where $\displaystyle sin^2(x)cos^2(x)=1$

Thanks for your help.