How do you solve #sinxcosx=1/2#?

1 Answer

Dec 27, 2016

#x= pi/4 + kpi#

Explanation:

#sin x .cos x = 1/2#
Replace in the equation (sin x.cos x) by #(sin 2x)/2# -->
#(sin 2x)/2 = 1/2#
#sin 2x = 1#
Trig unit circle -->
#2x = pi/2 + 2kpi#
#x = pi/4 + kpi#

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How do you solve #sinxcosx=1/2#?

1 Answer

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