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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