How do you solve for real values of x with the equation 2 cos² x = 3 sin x ?

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Answer 1 · 2018-02-22T20:08:39

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$2 cos^2x - 3sin x = 0$

$2(1 - sin^2 x) - 3sin x = 0$

$2 - 2sin^2 x - 3sin x = 0$

$2 sin^2 x + 3sin x - 2 = 0$

$( 2 sin x - 1)(sin x + 2) = 0$

So

$sin x = 1/2" "$ or $" "sin x = - 2$

where $-2$ is not in $[-1,1]$ . Using the formula for $sin x$ , we get

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