How do you solve #\sin (3x-27)=\cos (5x+5)#?

1 Answer
Nghi N
Jan 25, 2018

#x = 14^@ + k360^@#
#x = - 61^@ + k360^@#

Explanation:

Re-write the equation:
#cos (5x + 5^@) = sin (3x - 27^@)#
Use trig identity:
#sin a = cos (90^@ - a)#
In this case:
#sin (3x - 27) = cos (90 - 3x + 27) = cos (117 - 3x)#
We get from the definition of the cosine function:
#cos (5x + 5^@) = cos (117 - 3x)# -->
#(5x + 5) = +- (117 - 3x)#
a. 5x + 5 = 117 - 3x -->
8x = 112 -->
#x = 112/8 = 14^@ + k360^@#.
b. 5x + 5 = - 117 + 3x
2x = - 122 -->
#x = - 122/2 = - 61^@ + k360^@ #

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