SOLUTION: Find all solutions of the equation in the interval [0, 2π). 2 sin x/2 + 2 cos x = 0 2 cos x/2 − 2 sin x = 0

Algebra ->  Trigonometry-basics -> SOLUTION: Find all solutions of the equation in the interval [0, 2π). 2 sin x/2 + 2 cos x = 0 2 cos x/2 − 2 sin x = 0      Log On


   

Question 850228: Find all solutions of the equation in the interval [0, 2π).
2 sin x/2 + 2 cos x = 0
2 cos x/2 − 2 sin x = 0
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find all solutions of the equation in the interval [0, 2π;).
2 sin x/2 + 2 cos x = 0
2 cos x/2 -2 sin x = 0
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2 sin x/2 + 2 cos x = 0
2%28sin%28x%2F2%29%2Bcos%28x%29%29=0
%28sin%28x%2F2%29%2Bcos%28x%29%29=0
sqrt%28%281-cos%28x%29%29%2F2%29%2Bcos%28x%29=0
sqrt%28%281-cos%28x%29%29%2F2%29=-cos%28x%29
square both sides:
%281-cos%28x%29%29%2F2=cos%5E2%28x%29
2cos%5E2%28x%29%2Bcos%28x%29-1=0
(2cosx-1)(cosx+1)=0
cosx=1/2
x=π/3, 5π/3
or
cosx=-1
x=π
..
2 cos x/2 -2 sin x = 0
2%28cos%28x%2F2%29-sin%28x%29%29=0
%28cos%28x%2F2%29-sin%28x%29%29=0
sqrt%28%281%2Bcos%28x%29%29%2F2%29-sqrt%281-cos%5E2%28x%29%29=0
sqrt%28%281%2Bcos%28x%29%29%2F2%29=sqrt%281-cos%5E2%28x%29%29
square both sides
%281%2Bcos%28x%29%29%2F2=1-cos%5E2%28x%29%29
1+cosx=2-2cos^2x
2cos^2x+cosx-1=0
(same answers as first equation)
x=π/3, 5π/3,π