How do you solve #(sin x)^2 + 5 cos x +5=0#?

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How do you solve #(sin x)^2 + 5 cos x +5=0#?

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Answer 1 · 2016-04-27T18:02:43

#pi + 2kpi#

Explanation:

Replace in the equation #sin^2 x# by #(1 - cos^2 x) #-->
#1 - cos^2 x + 5cos x + 5 = 0#
#-cos^2 x + 5cos x + 6 = 0. #
Solve this quadratic equation for cos x.
Since a - b + c = 0, the 2 real roots are:
#cos x = -1 # and #cos x = -c/a = -6/-1 = 6# (rejected as > 1)
cos x = -1 --> #x = pi#
General answers: #x = pi + 2kpi#

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How do you solve #(sin x)^2 + 5 cos x +5=0#?

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