
Solve [0,2pi)
2 similar problems that are really challenging for me.
1) solve [0,2pi) sin 2xsinx = 0
2) solve [0,2pi) tan^2x sinx= tan^2x
Also how do i rewrite cos^3 x so that it does not contain power greater than 1? I don't think i can use double angle identity since it has power of 3.. not multiples of 2
(Headbang)
Thanks!!

Re: Solve [0,2pi)
$\displaystyle \sin 2x  \sin x = 0 $
use the identity $\displaystyle \sin 2x = 2\sin x \cos x $ , then take as a factor sinx
$\displaystyle \sin x (2 \cos x  1 ) = 0 $
$\displaystyle \sin x = 0 \Rightarrow x = 0,\pi,2\pi $
$\displaystyle 2\cos x  1 = 0 \Rightarrow \cos x = \frac{1}{2}\;\Rightarrow x = \frac{\pi}{3},\frac{5\pi}{3} $
second one move the left side to the right side then take as a factor tan^2 x same strategy above