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Math Help - Solve trig equatiom

  1. #1
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    Solve trig equatiom

    The equation is on the interval [0,2pi]:2cos^2x+3sinx-3=0

    I get up to the part -2sinx^2+3sinx-1=0

    I cant get past this, im stuck as to how i should factor it.

    Can domeone also tell me if i can do this on the ti-nspire cas? If so, how?
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  2. #2
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    Re: Solve trig equatiom

    2\cos^2 x+3\sin x-3=0

    2(1-\sin^2 x)+3\sin x-3=0

    2-2\sin^2 x+3\sin x-3=0

    -2\sin^2 x+3\sin x-1=0

    make \sin x = u

    -2u^2 +3u-1=0

    factor the quadratic
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  3. #3
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    Re: Solve trig equatiom

    Ok i had that part....i factored it as

    (-sinx+1)(2sinx-1)=0

    Sin x =1 which is pi/2 and 2sinx=0 which is 2pi?

    Can someone please confirm?
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    Re: Solve trig equatiom

    The second factor

    2\sin x-1 = 0

    2\sin x = 1

    \sin x = \frac{1}{2}

    x = \frac{\pi}{6}
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  5. #5
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    Re: Solve trig equatiom

    Whoops...thanks for pointing that out!
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  6. #6
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    Re: Solve trig equatiom

    Wait, somethings still wrong...

    The first equation is

    -sinx-1=0
    And that gives me
    -sinx=1 and as per the even and odd identity it is the same as sinx=-1 correct?
    So what would my ans be? 3pi/2?

    And for the second equation, sin x =1/2 so would my ans be pi/6+2npi and 5pi/6+2npi?
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  7. #7
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    Re: Solve trig equatiom

    I think i still have the factoring wrong, can someone please help?
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  8. #8
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    Re: Solve trig equatiom

    The equation is on the interval [0,2pi]:2cos^2x+3sinx-3=0
    -2\sin^2{x} + 3\sin{x} - 1 = 0

    2\sin^2{x} - 3\sin{x} + 1 = 0

    (2\sin{x}-1)(\sin{x}-1) = 0

    \sin{x} = \frac{1}{2}

    \sin{x} = 1

    now, pay attention to the given interval in the original problem statement and write the three solutions for x
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  9. #9
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    Re: Solve trig equatiom

    I see, why do i have to multiply by neg -1?

    And the value for sin x=1 is pi/2+2npi? And for sin 1/2 is pi/6+2npi and 5pi/6+2npi?

    I have to write the 2npi, right?
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  10. #10
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    Re: Solve trig equatiom

    Quote Originally Posted by noork85 View Post
    I see, why do i have to multiply by neg -1? makes it easier to factor

    And the value for sin x=1 is pi/2+2npi? And for sin 1/2 is pi/6+2npi and 5pi/6+2npi?

    I have to write the 2npi, right?
    NO, read this again ...

    The equation is on the interval [0,2pi]:2cos^2x+3sinx-3=0
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  11. #11
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    Re: Solve trig equatiom

    I thought that because its on the 2pi interval, it has infinitely more solutions, which is why its 5pi/6 +2npi and so on...

    If not that, then i really dont know...
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