How do you solve #cos((3theta)/5 + 11)=sin ((7theta)/10 + 40s)#?

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Answer 1 · 2017-05-17T00:06:44

#s=((13theta)/10-101)/-40#

Use the cofunction rule:

#cos(x)=sin(90-x)#
#sin(x)=cos(90-x)#.

It doesn't really matter which one you want to convert. I will convert cos to sin. Set #x# equal to #(3theta)/5+11#

#cos((3theta)/5+11)=sin(90-(3theta)/5+11)#

So now we have:
#sin(90-(3theta)/5+11)=sin((7theta)/10+40s)#

Now, inverse #sin# both sides. This will cancel out the #sin#:

#90-(3theta)/5+11=(7theta)/10+40s#

Simplify so that our problem looks a little nicer:

#101-(3theta)/5=(7theta)/10+40s#

Move #(3theta)/5# and #40s# over to the other side:

#101-40s=(13theta)/10#

Solve for #s#:

#-40s=(13theta)/10-101#

#s=((13theta)/10-101)/-40#