SOLUTION: Find all solutions on the interval [0,2pi) for the equation: 2sin^2(x)=cos(x)+1

Algebra ->  Trigonometry-basics -> SOLUTION: Find all solutions on the interval [0,2pi) for the equation: 2sin^2(x)=cos(x)+1      Log On


   

Question 330582: Find all solutions on the interval [0,2pi) for the equation: 2sin^2(x)=cos(x)+1
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
2sin%5E2%28x%29=cos%28x%29%2B1
2%281-cos%5E2%28x%29%29=cos%28x%29%2B1
2-2cos%5E2%28x%29-cos%281%29-1=0
-2cos%5E2%28x%29-cos%281%29%2B1=0
2cos%5E2%28x%29%2Bcos%281%29-1=0
%282cos%28x%29-1%29%28cos%28x%29%2B1%29=0
Solve both:
2cos%28x%29-1=0
cos%28x%29=1%2F2
x=60degrees and x=300degrees
.
.
.
cos%28x%29%2B1=0
+cos%28x%29=-1+
x=180degrees
.
.
.
Three solutions: 60,180,300 degrees (pi%2F3,pi,5pi%2F3) radians