Question

How do you solve `cos^2x sinx =0` ?

Answer 1

In [ 0, 2`pi` ], the solutions are 0, `pi`/2#`,`pi`/, 3`pi`/2 and 2`pi#.

Explanation:

The equation is sin x ( 1 `-` sin x , sin x) = 0.
sin x = 0 gives solutions 0, `pi` and 2`pi`.
The other factor = 0 breaks into sin x = `+-`1, giving the other two solutions.