How do you solve $sin^2xcosx=cosx$ ?

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How do you solve $sin^2xcosx=cosx$ ?

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Answer 1 · 2016-08-25T00:48:31

$pi/2,, (3pi)/2$

Explanation:

Bring the equation to standard form:
$sin^2 x.cos x - cos x = 0$
$cos x(sin^2 x - 1) = 0$

a. cos x = 0 --> Unit circle -->
$x = pi/2$ , and $x = (3pi)/2$
b. $sin^2 x - 1 = 0$ --> $sin^2 x = 1$
$sin x = +- 1$
1. sin x = - 1 --> $x = (3pi)/2$
2. sin x = 1 --> $x = pi/2$

Answers for $(0, 2pi)$ :
$pi/2, (3pi)/2,$

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