How do you solve for x if #cos (6x-20) = sin (2x - 10)#?

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Answer 1 · 2018-07-24T02:30:19

#x=15#

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

#cos (6x-20)=sin(90-(6x-20))=sin(90-6x+20)=sin(110-6x)#

#sin(110-6x)=sin(2x-10)#

#110-6x=2x-10#

#120=8x#

#x=15#

Sub #x=15# into the equation
LHS:
#cos(6x-20)#
#=cos(6times15-20)#
#=cos(90-20)#
#=cos70#

RHS:
#sin(2x-10)#
#=sin(2times15-10)#
#=sin(30-10)#
#=sin20#
#=cos(90-20)#
#=cos70#
#=LHS#

Therefore, #x=15# is correct