How do you solve # sin(x+pi) + cos(x+pi) = 0# from 0 to 2pi?

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Answer 1 · 2016-02-02T08:13:55

#sin(x+pi)=-sin(x)=sin(-x)=cos(pi/2 +x)#

#cos (x+pi)=-cos(x)#

So According to the question #cos(pi/2 +x)=cos(x)#

Using solution of cos we get

#(pi/2) + x = 2npi +- (x)#

So case I:- #(pi/2) + x = 2npi + (x)#

But here, the value of x is cancelled, so

case II:- #(pi/2) + x = 2npi - (x)#

#x=npi-pi/4#

The solution of this equation is #x=npi-pi/4#