How do you solve #4sinxcosx=1#?

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How do you solve #4sinxcosx=1#?

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Answer 1 · 2015-07-06T03:35:01

#x=pi/12+kp#i,
#x = (5pi)/12+kpi#

Explanation:

Use the trig identity:
#sin 2x = 2sin xcos x#
In this case,
#4sin xcos x = 2sin 2x = 1# =>
#sin 2x = 1/2#
Trig table and unit circle give 2 solutions:
a. #2x=pi/6 + 2kpi# -->
#x=pi/12 + kpi#.
b. #2x=(5pi)/6 + 2kpi#, -->
#x=(5pi)/12 + kpi#.

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How do you solve #4sinxcosx=1#?

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