SOLUTION: Please help me with this problem. I need to solve the equation for solutions over the interval [0 degrees,360 degrees): {{{ cos^2(x)=sin^2(x)+1 }}}

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me with this problem. I need to solve the equation for solutions over the interval [0 degrees,360 degrees): {{{ cos^2(x)=sin^2(x)+1 }}}      Log On


   

Question 494838: Please help me with this problem. I need to solve the equation for solutions over the interval [0 degrees,360 degrees):
+cos%5E2%28x%29=sin%5E2%28x%29%2B1+
Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
Substitute in our equation %28sinx%29%5E2=1-%28cosx%29%5E2
%28cosx%29%5E2=%281-%28cox%29%5E2%29%2B1 => 2%28cosx%29%5E2=2 => %28cosx%29%5E2=1
From the final equation we get: cosx=1 and cosx=-1. Solving this two equations we
get:1) cosx=1 => x=2n%2Api, and 2)cosx=-1 => x=%282n-1%29%2Api, where
n=0, +-1, +-2, ...