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

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

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Answer 1 · 2016-02-23T12:31:45

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.

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

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