SOLUTION: Solve {{{ cos(2x)-sin(x)=0 }}} given that 0 ≤ x < 2π
--Not sure what to select for the topic of this question, and I am not sure what the process is to solve it. I wo
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Question 1098062: Solve given that 0 ≤ x < 2π
--Not sure what to select for the topic of this question, and I am not sure what the process is to solve it. I would appreciate any pointers, thank you!
Found 3 solutions by CubeyThePenguin, ikleyn, ewatrrr:
Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
cos(2x) = 1 - 2sin^2(x)
Rewrite as a quadratic in terms of sin(x). Let sin(x) = y if you like.
1 - 2sin^2(x) - sin(x) = 0
-2y^2 - y + 1 = 0
2y^2 + y - 1 = 0
(2y - 1)(y + 1) = 0
sin(x) = 1/2 --> x = pi/6
sin(x) = -1 --> x = 3pi/2
Answer by ikleyn(52847) (Show Source): You can put this solution on YOUR website!
.
In the post by @CubeyThePenguin, one root is missed and should be added.
This root is .
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
sin(x) = 1/2 --> x = pi/6 and x = 5pi/6
sin(x) = -1 --> x = 3pi/2
(cos(x), sin(x))
.
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