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Math Help - Angle Question

  1. #1
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    Angle Question

    So I have two angle problems...

    sin2x = 2cosxcos2x ; which becomes
    0 = 2cosxcos2x - sin2x
    0 = 2cosxcos2x - 2sinxcosx
    0 = 2cosx(cos2x - sin x)

    so either

    2cosx = 0 OR cos2x - sinx = 0

    Thank you.
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  2. #2
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    Quote Originally Posted by CrimesofParis
    So I have two angle problems...

    sin2x = 2cosxcos2x ; which becomes
    0 = 2cosxcos2x - sin2x
    0 = 2cosxcos2x - 2sinxcosx
    0 = 2cosx(cos2x - sin x)

    so either

    2cosx = 0 OR cos2x - sinx = 0

    Thank you.

    Okay you did everything correctly up till now. Now you are required to solved for x.
    The first one 2\cos x=0. Divide by two,
    \cos x=0, now you need to know when \cos x is zero. That is when x=90,270 as two non-coterminal angels. To find all solutions it is the same as x=90+360k and x=270+360k=-90+360+360k=-90+360(1+k) where k is any integer. Because as you know that addition and subtraction of 360 keeps the trigonometric function of that angel the same. But k+1 is also an integer. Thus, all solutions for x have form. x=(-1)^{k+1}90+360k, k \epsilon Z or in radian form x=(-1)^{k+1}\frac{\pi}{2}+2\pi k,k \epsilon Z
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  3. #3
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    For your second question:
    \cos 2x-\sin x=0
    Use the identity
    1-2\sin^2x=\cos 2x
    Thus,
    1-2\sin^2x-\sin x=0
    Change signs,
    2\sin^2x+\sin x-1=0
    Factor,
    (2\sin x-1)(\sin x +1)=0
    Now each one is set to zero,
    2\sin x-1=0
    Thus,
    \sin x=\frac{1}{2}
    That is when x=30,150 as the only two non-coterminal angels. Thus all solution have form x=30+360k, 150+360k But,
    x=30+360k=30+180(2k)
    x=150+360k=-30+180(2k+1)
    Thus, all solutions have form x=(-1)^k30+180k or in radian form x=(-1)^k\frac{\pi}{6}+\pi k,k \epsilon Z
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  4. #4
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    Finally the last factor equal to zero, (of the quadratic)
    \sin x+1=0
    Thus,
    \sin x=-1 That happens when x=270 and that is the only non-coterminal angle. Thus all solutions have form x=270+360k. But 270+360k=-90+360+360k=-90+360(k+1).
    Thus, all solutions are x=-90+360k. Or in radian form x=-\frac{\pi}{2}+2\pi k,k\epsilon Z.

    If you have difficulty understanding all that manipulating after the solutions just understand that by adding 360 or subtracting 360 you change nothing. Thus, those are all the solutions. All I did was manipulated into a more elegant form.
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  5. #5
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    Thank you so much, that pretty much answers my questions, except I forgot to mention the answer has to be radians, so I should be able to covert the middle steps to radians. But thank you.
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