Expert: Sherman D. - 11/2/2008

Question
hey. so im doing this problem but i'm kind of stuck. it says:
solve: (1/3)cos²x = sin²x, for 0<x<2(pi)

Answer
There are 2 ways to do this

(1/3)cos(x)^2 = sin(x)^2
(1/3)(1 - sin(x)^2) = sin(x)^2
1 - sin(x)^2 = 3sin(x)^2
4sin(x)^2 = 1
sin(x)^2 = 1/4
sin(x) = (1/2)
x = ±1/2

2sin(x)^2 - 1 = 0
2sin(x)^2 = 1
sin(x)^2 = 1/4
sin(x) = (1/2)

x = 30°, 150° for the positive of (1/2)
x = 210°, 330° for the negative of (1/2)

x = 30°, 150°, 210°, and 330°

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(1/3)cos(x)^2 = sin(x)^2
(1/3)cos(x)^2 = 1 - cos(x)^2
cos(x)^2 = 3 - 3cos(x)^2
4cos(x)^2 = 3
cos(x)^2 = (3/4)
cos(x) = sqrt(3)/2

x = 30°, 330° for the positive of sqrt(3)/2
x = 150°, 210° for the negative of sqrt(3)/2

x = 30°, 150°, 210°, and 330°