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Math Help - Trig

  1. #1
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    Trig

    Solve the following for x: 2sin2(x)=1​, 0x2

    The solution set is _______ For example if there are 3 solutions, then write {0, pi/3, 3pi/4}

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  2. #2
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    Re: Trig

    Quote Originally Posted by bookish12 View Post
    Solve the following for x: 2sin2(x)=1​, 0x2

    The solution set is _______ For example if there are 3 solutions, then write {0, pi/3, 3pi/4}
    2\sin(2x)=1\Rightarrow \sin(2x)=\frac{1}{2}\Rightarrow 2x=\frac{\pi}{6}+2k\pi \text{ or } 2x=\frac{5\pi}{6}+2k\pi\Rightarrow\\ \Rightarrow x=\frac{\pi}{12}+k\pi \text{ or } x=\frac{5\pi}{12}+k\pi, k\in\mathbb{Z}.

    Since 0\leq x\leq 2\pi \Leftrightarrow  x\in[0,2\pi]=\left\[ 0, \frac{24\pi}{12}  \right\] you just have to check for what values of k\in\mathbb{Z} solutions given above fall into the interval of interest.
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  3. #3
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    Re: Trig

    Hello, bookish12!

    I assume that the second "2" is an exponent.


    \text{Solve: } 2\sin^2(x) \:=\: 1,\;\;0 \le x < 2\pi

    \begin{array}{c}2\sin^2(x) \:=\:1 \\ \\ \sin^2(x) \:=\:\frac{1}{2} \\ \\ \sin(x) \:=\:\pm\dfrac{1}{\sqrt{2}} \\ \\ x \;=\;\dfrac{\pi}{4},\,\dfrac{3\pi}{4},\,\dfrac{ 5\pi}{4},\,\dfrac{7\pi}{4} \end{array}
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  4. #4
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    Re: Trig

    Thank you!!
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