Solve: sin x + 2 sin² x = 0 when (0˚ ≤ x < 360˚)
I really suck at these problems... I just have trouble remembering the unit circle. Any help (or the answer of course) would be appreciated. And if you're up for it, could you try explaining the process to me? Usually there are a couple different answers for these types of problems btw. Thanks.
I really suck at these problems... I just have trouble remembering the unit circle. Any help (or the answer of course) would be appreciated. And if you're up for it, could you try explaining the process to me? Usually there are a couple different answers for these types of problems btw. Thanks.
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sin(x) + 2sin^2(x) = 0
Factor out our common term (sin(x)):
sin(x) * (1 + 2sin(x)) = 0
Thus, we have two solutions:
sin(x) = 0 and 1 + 2sin(x) = 0
We need to manipulate this second one a bit to solve for sin(x). First subtract 1:
2sin(x) = -1
Divide by 2:
sin(x) = -1/2
Thus, we have solutions when sin(x) = 0 and when sin(x) = -1/2
This, by any method you choose will give solutions:
x = 0 and x = 180 for the sin(x) = 0 piece and x = 210 and x = 330 for the second piece.
So, we have four solutions: x = 0, x = 180, x = 210, and x = 330 degrees.
Factor out our common term (sin(x)):
sin(x) * (1 + 2sin(x)) = 0
Thus, we have two solutions:
sin(x) = 0 and 1 + 2sin(x) = 0
We need to manipulate this second one a bit to solve for sin(x). First subtract 1:
2sin(x) = -1
Divide by 2:
sin(x) = -1/2
Thus, we have solutions when sin(x) = 0 and when sin(x) = -1/2
This, by any method you choose will give solutions:
x = 0 and x = 180 for the sin(x) = 0 piece and x = 210 and x = 330 for the second piece.
So, we have four solutions: x = 0, x = 180, x = 210, and x = 330 degrees.