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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