Question 494838
Substitute in our equation {{{(sinx)^2=1-(cosx)^2}}}

{{{(cosx)^2=(1-(cox)^2)+1}}} => {{{2(cosx)^2=2}}} => {{{(cosx)^2=1}}}

From the final equation we get: cosx=1 and cosx=-1. Solving this two equations we 

get:1) cosx=1 => {{{x=2n*pi}}}, and 2)cosx=-1 => {{{x=(2n-1)*pi}}}, where

n=0, +-1, +-2, ...