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Thread: Please help with solving trig equations!

  1. #1
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    Please help with solving trig equations!




    THANKS!
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  2. #2
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    Quote Originally Posted by gobbajeezalus View Post



    THANKS!
    (1) $\displaystyle \sin x \tan x - \sin x=0$

    $\displaystyle \sin x(\tan x-1)=0$

    $\displaystyle \sin x = 0 \ \ or \ \ \tan x-1=0$

    $\displaystyle \tan x=1$

    $\displaystyle \sin x=0 \ \ at \ \ \left\{0, \pi \right\}$

    $\displaystyle \tan x=1 \ \ at \ \ \left\{\frac{\pi}{4}, \frac{5\pi}{4}\right\}$

    Solution set=$\displaystyle \left\{0, \frac{\pi}{4}, \pi, \frac{5\pi}{4}\right\}$
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  3. #3
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    Quote Originally Posted by gobbajeezalus View Post



    THANKS!
    (2) $\displaystyle \sin 2x=-\sin x$

    $\displaystyle \sin 2x+\sin x$

    Double angle identity: $\displaystyle \sin 2x = 2\sin x \cos x$

    $\displaystyle 2\sin x \cos x + \sin x=0$

    $\displaystyle \sin x(2\cos x+1)=0$

    $\displaystyle \boxed{\sin x = 0} \ \ or \ \ 2\cos x+1=0$

    $\displaystyle 2\cos x=-1$

    $\displaystyle \boxed{\cos x = -\frac{1}{2}}$

    $\displaystyle \sin x=0 \ \ at \ \ \left\{0, \pi\right\}$

    $\displaystyle \cos x = -\frac{1}{2} \ \ at \ \ \left\{\frac{2\pi}{3}, \frac{4\pi}{3}\right\}$

    Solution set=$\displaystyle \left\{0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3}\right\}$
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  4. #4
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    Quote Originally Posted by gobbajeezalus View Post



    THANKS!
    (3) $\displaystyle 2\sin \theta \cos \theta + \sqrt{3} \sin \theta = 0$

    $\displaystyle \sin \theta(2 \cos \theta + \sqrt{3})=0$

    $\displaystyle \boxed{\sin \theta=0} \ \ or \ \ 2 \cos \theta = -\sqrt{3}$

    $\displaystyle \boxed{\cos \theta = -\frac{\sqrt{3}}{2}}$

    $\displaystyle \sin \theta = 0 \ \ at \ \ \left\{0, \pi\right\}$

    $\displaystyle \cos \theta = -\frac{\sqrt{3}}{2} \ \ at \ \ \left\{\frac{5\pi}{6}, \frac{7\pi}{6}\right\}$

    Solution set=$\displaystyle \left\{0, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}\right\}$
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