How do you solve #cosx + 1 = sinx#?
1 Answer
Nov 21, 2017
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
Hopefully this helps!