How do you solve #cosx=sinx#?

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How do you solve #cosx=sinx#?

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Answer 1 · 2016-08-09T15:57:23

#pi/4 or 45^@#

Explanation:

Property of complementary arcs -->
#cos x = sin (pi/2 - x)#
There for,
#sin x = sin (pi/2 - x)#
a. #x = (pi/2 - x)# --> #2x = pi/2# --> #x = pi/4#
b. #x = pi - (pi/2 - x) = pi/2 + x# (Rejected as non-sense)

Answer: #x = pi/4 or 45^@#
Check.
#x = pi/4# --> #sin x = sqrt2/2# --> #cos x = sqrt2/2# . OK

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How do you solve #cosx=sinx#?

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