How do you solve #sin(2x)=-cos(2x)#?

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How do you solve #sin(2x)=-cos(2x)#?

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Answer 1 · 2017-06-30T18:05:45

#x = (3pi)/8 + kpi#
#x = (7pi)/8 + kpi#

Explanation:

sin 2x + cos 2x = 0
Use trig identity: #sin a + cos a = sqrt2cos (a - pi/4)#
In this case:
#sin 2x + cos 2x = sqrt2cos (2x - pi/4) = 0#
#cos (2x - pi/4) = 0#
Unit circle gives 2 solutions:
a. #2x - pi/4) = pi/2 + 2kpi# --> #2x = pi/2 + pi/4 = (3pi)/4 + 2kpi# -->
#x = (3pi)/8 + kpi#.
b. #(2x - pi/4) = (3pi)/2 + 2kpi# --> #2x = (3pi)/2 + pi/4 = (7pi)/4 + 2kpi# -->#x = (7pi)/8 + kpi#

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How do you solve #sin(2x)=-cos(2x)#?

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