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Thread: some more trig identities

  1. #1
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    some more trig identities

    hi. I think i'm getting the hang of these problems but I still need some assistance:

    1) $\displaystyle cos^2x+sinx=1$
    $\displaystyle cos^2x+sinx=cos^2x+sin^2x \rightarrow sinx - sin^2x=0 \rightarrow sinx(1-sinx)=0$
    $\displaystyle sinx_1 = 0 \rightarrow x_1=180n$
    $\displaystyle 1-sinx_2=0 \rightarrow sinx_2=1 \rightarrow x_2=90+360n$

    2) $\displaystyle 4sin^3x-sinx=0$
    $\displaystyle sinx(4sin^2x-1)=0$
    $\displaystyle sinx_1 = 180n$
    $\displaystyle 4sin^2x_2-1=0 \rightarrow sin^2x=1/4 \rightarrow sinx=1/2 \rightarrow x_2 = +-30+180n$

    3) $\displaystyle 2cos^2x-5sinx-4=0$
    $\displaystyle 2-2sin^2x -5sinx-4=0$ ???

    are 1 and 2 correct? number 2 I wasn't sure about (sinx)^2=1/4 - - - sinx=1/2. can i just do a square root like that? and how do I solve 3?

    Thanks
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  2. #2
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    Quote Originally Posted by jayshizwiz View Post
    hi. I think i'm getting the hang of these problems but I still need some assistance:

    1) $\displaystyle cos^2x+sinx=1$
    $\displaystyle cos^2x+sinx=cos^2x+sin^2x \rightarrow sinx - sin^2x=0 \rightarrow sinx(1-sinx)=0$
    $\displaystyle sinx_1 = 0 \rightarrow x_1=180n$
    $\displaystyle 1-sinx_2=0 \rightarrow sinx_2=1 \rightarrow x_2=90+360n$

    2) $\displaystyle 4sin^3x-sinx=0$
    $\displaystyle sinx(4sin^2x-1)=0$
    $\displaystyle sinx_1 = 180n$
    $\displaystyle 4sin^2x_2-1=0 \rightarrow sin^2x=1/4 \rightarrow sinx=1/2 \rightarrow x_2 = +-30+180n$

    3) $\displaystyle 2cos^2x-5sinx-4=0$
    $\displaystyle 2-2sin^2x -5sinx-4=0$ ???

    are 1 and 2 correct? number 2 I wasn't sure about (sinx)^2=1/4 - - - sinx=1/2. can i just do a square root like that? and how do I solve 3?

    Thanks
    (1) is good...

    For (2), you neglected to write the negative square root of 1/4, which is -1/2, but you found the angles nonetheless.

    (3) can written as a quadratic equation, using

    $\displaystyle -2Sin^2x-5Sinx-2=0$

    $\displaystyle 2Sin^2x+5Sinx+2=0$

    $\displaystyle (2Sinx+1)(Sinx+2)=0$

    and work from there
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  3. #3
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    thanks. i can't believe i didn't realize that. here's what I did...



    $\displaystyle sinx(2sinx+5)=-2$ and then i was stuck...lol
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