How do you solve for X in cos(x+30)=sin(5x+12)?

Question

I dont get it, like i get trig and other stuff, but i dont know how to begin solving this problem

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Answer 1 · 2017-01-24T18:09:51

Please see the explanation.

There are two base angles that solve this, #8^@ and 27^@#.

The first angle is found by converting the cosine function into the negative of the sine function by subtracting #90^@#:

#-sin(x - 60^@) = sin(5x + 12^@)#

Use the identity #-sin(a) = sin(-a)#

#sin(-x + 60^@) = sin(5x + 12^@)#

Equate the arguments:

#-x + 60^@ = 5x + 12^@#

#6x = 48^@#

#x = 8^@#

Because #5x - -x = 6x#, this repeats 6 times every cycle or every #60^@#

#x = 8^@ + n60^@# where n is any integer positive or negative including zero.

The second angle is found by converting the sine function into the cosine function by subtracting #90^@#:

#cos(x + 30^@) = cos(5x - 78^@)#

Equate the arguments:

#x + 30^@ = 5x - 78^@#

#4x = 128^@#

#x = 27^@#

Because #5x - x = 4x# this repeats 4 times every cycle or every #90^@#

#x = 27^@ + n90^@#

The complete answer that will give you all of the solutions is:

#x ={(8^@ + n60^@),(27^@+n90^@):}# where n is any positive or negative integer, including 0.