How do you solve #sin^2x+ 3/2cos(2x) = 0#?

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How do you solve #sin^2x+ 3/2cos(2x) = 0#?

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Answer 1 · 2018-04-22T03:39:46

Use relevant identities.

Explanation:

#sin^2(x) + 3/2cos(2x) = 0# = #sin^2(x) + 3/2(1-2sin^2(x))= 0#
Simplifying will get you
#-2sin^2(x) = -3/2#
#sin^2(x) = 3/4#
#sin(x) = sqrt(3)/2# and #sin(x) = -sqrt(3)/2#
Basic Angle = #sin^-1(sqrt(3)/2)#
Basic Angle = #60^.#
Since #sin# is both positive and negative, angle must lie in all quadrants.
Thus #angle# = #60, 180^(.)-60^(.), 180^(.) + 60^(.), 360^(.) - 60^(.)#

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How do you solve #sin^2x+ 3/2cos(2x) = 0#?

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