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  1. #1
    Junior Member Lonehwolf's Avatar
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    Show that...

    Show that the equation 15 cos x = 13+sin x may be written as a quadratic equation in sin x.

    Hence solve the equation, giving all values of "x" such that 0<=x<=360.
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Lonehwolf View Post
    Show that the equation 15 cos x = 13+sin x may be written as a quadratic equation in sin x.

    Hence solve the equation, giving all values of "x" such that 0<=x<=360.
    Square both sides, and use the Pythagorean identity, \sin^2x+\cos^2x=1. Collect terms and solve.
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  3. #3
    Junior Member Lonehwolf's Avatar
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    Didn't help me, this is what I did:

    15 cos^2x-(13+sin x)^2
    225cos^4x=(13+sin x)(13+sin x)
    225cos^4x=196+26sin x +sin^2x

    Was that actually going good and I simply didn't notice or did I get what you meant with square both sides wrong?
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  4. #4
    Moo
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    A Cute Angle Moo's Avatar
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    Quote Originally Posted by Lonehwolf View Post
    Didn't help me, this is what I did:

    {\color{red}15} cos^2x-(13+sin x)^2
    225cos^{\color{red}4}x=(13+sin x)(13+sin x)
    225cos^{\color{red}4}x=196+26sin x +sin^2x

    Was that actually going good and I simply didn't notice or did I get what you meant with square both sides wrong?
    Why the red things ?

    \begin{aligned}<br />
196+26 \sin(x) +\sin^2(x)<br />
&=(15 \cos(x))^2 \\<br />
&=225 \cos^2(x) \\<br />
&=225-225\sin^2(x) \end{aligned}

    and this gives your quadratic
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  5. #5
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Lonehwolf View Post
    Didn't help me, this is what I did:

    15 cos^2x-(13+sin x)^2
    225cos^4x=(13+sin x)(13+sin x)
    225cos^4x=196+26sin x +sin^2x

    Was that actually going good and I simply didn't notice or did I get what you meant with square both sides wrong?
    I am not sure what you are doing in the above. How did squaring cosine produce \cos^4x? Is the equation in the original post the correct equation?

    You should have

    15\cos x=13+\sin x

    \Rightarrow225\cos^2x=\sin^2x+26\sin x+169

    \Rightarrow225\left(1-\sin^2x\right)=\sin^2x+26\sin x+169

    \Rightarrow225-225\sin^2x=\sin^2x+26\sin x+169

    \Rightarrow-226\sin^2x-26\sin x+56=0.

    There is your quadratic, now solve it.

    Quote Originally Posted by Moo View Post
    Why the red things ?

    \color{red}196\color{black}+26 \sin(x) +\sin^2(x)=\dots
    Yes, Moo. Why the red things?
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  6. #6
    Junior Member Lonehwolf's Avatar
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    Ah shame on me wrong first question -.-

    it should be

    15cos^2x=13+sinx
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  7. #7
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    Quote Originally Posted by Lonehwolf View Post
    Ah shame on me wrong first question -.-

    it should be

    15cos^2x=13+sinx
    Then replace cos^2(x) with 1- sin^2(x). This is far easier than the problem you originally posted!
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