How do you solve #4sin(x+pi)=2cos(x+pi/2)+2#?

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Answer 1 · 2016-09-14T09:27:02

#x=2npi-pi/2#

We can solve the equation #4sin(x+pi)=2cos(x+pi/2)+2#,

by using the identities #sin(x+pi)=-sinx# and #cos(x+pi/2)=-sinx#

Hence #4sin(x+pi)=2cos(x+pi/2)+2# can be written as

#-4sinx=-2sinx+2# or

#-4sinx+2sinx=2# or

#-2sinx=2# or

#sinx=-1=son(-pi/2)#

Hence #x=2npi-pi/2#