How do you solve #cosx + 1 = sinx#?

1 Answer
Noah G
Nov 21, 2017

#x = pi/2 and pi#

Explanation:

I'm assuming you want us to solve the equation.

#(cosx+ 1)^2 = (sinx)^2#

#cos^2x + 2cosx + 1 = sin^2x#

#cos^2x + 2cosx + 1 - sin^2x = 0#

#cos^2x + 2cosx + 1 -(1 - cos^2x) = 0#

#cos^2x + 2cosx + 1 - 1 + cos^2x = 0#

#2cos^2x + 2cosx= 0#

#2cosx(cosx + 1) = 0#

This means that #cosx = 0# or #cosx = -1#. Thus, #x = pi/2, pi, (3pi)/2#. But #x = (3pi)/2# is extraneous since it doesn't satisfy the initial equation.

Hopefully this helps!

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