How do you solve #sinx+cosx=0#?

1 Answer

Apr 27, 2018

#(3pi)/4; (7pi)/4#

Explanation:

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

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How do you solve #sinx+cosx=0#?

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