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

1 Answer
Feb 2, 2016

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