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Math Help - Trigonometric equation

  1. #1
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    Trigonometric equation



    Work shown so far:





    This is where I'm stuck at. I know that I somehow have to get two terms to equal to zero, but I don't know how. The +1 at the end of the problem is throwing me for a loop. Am I even doing it correct?
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  2. #2
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    Correct up to \displaystyle 2 - 2\sin^2{x} + \sin{x} - 1 = 0.

    From here, you have forgotten that after that it is an EQUATION, and also there should be a negative on the \displaystyle 2\sin^2{x}.

    Solve it as you would a quadratic equation.
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  3. #3
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    Is this correct?



    Also if it is, how do you know when to add the +n(2pi) to each of the solutions? It asked to solve the equation on interval [0,2pi).
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  4. #4
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    I believe you mean \displaystyle \sin{x} = \frac{-1 \pm \sqrt{1 - 4(-2)(1)}}{2(-2)} = -\frac{1}{2} or \displaystyle 1...
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  5. #5
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    yessir that's what I mean hehe. Still stuck in algebra mode. apologies. Thanks again for ur help Prove It. Do you happen to know the answer to the question about the add on?
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  6. #6
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    Quote Originally Posted by isuckatallmath View Post
    By the identity 1+\cos{2x} = 2\cos^2{x}, we've 2\cos^2{x}+\sin{x}-1 = \cos{2x}+\sin{x}.
    So for your equation we have \cos{2x}+\sin{x} = 0 \Rightarrow \cos{2x} = -\sin{x} = \cos\left(x+\frac{\pi}{2}\right).
    If \cos{\theta} = \cos{\alpha}, then \theta = 2n\pi\pm \alpha, thus for the equation \cos{2x} = \cos\left(x+\frac{\pi}{2}\right) we've:
    2x = 2n\pi+x+\frac{\pi}{2} \Rightarrow x = 2n\pi+\frac{\pi}{2} or we've 2x = 2n\pi-x-\frac{\pi}{2} \Rightarrow x = \frac{2}{3}n\pi-\frac{\pi}{6}.
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