Question 870609: Find all solutions in the interval [0, 2pi) for the equation sin^2θ=cos^2θ-1
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! I'm going to use the variable x instead of the variable θ
Solve for x
sin^2(x)=cos^2(x)-1
sin^2(x)-cos^2(x) = -1
-1(-sin^2(x)+cos^2(x)) = -1
-sin^2(x)+cos^2(x) = 1
cos^2(x)-sin^2(x) = 1
cos(2x) = 1
2x = arccos(1)
2x = 0+2pi*n
2x = 2pi*n
x = 2pi*n/2
x = pi*n
All solutions for x are generated by x = pi*n where n is an integer
If n = -1, then
x = pi*n
x = pi*(-1)
x = -pi ... not in the interval [0, 2pi), so we ignore it
If n = 0, then
x = pi*n
x = pi*0
x = 0
If n = 1, then
x = pi*n
x = pi*1
x = pi
If n = 2, then
x = pi*n
x = pi*2
x = 2pi ... not in the interval [0, 2pi), so we ignore it
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The only two solutions are x = 0, x = pi (both in radians)
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