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Thread: trig equations

  1. #1
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    trig equations

    1. Which of the following is a solution to the equation 2sin^2x+sinx=1
    I. x=30
    II. x=150
    III. x=270

    a)III only
    b)I and III only
    c) II and III only
    d)I, II, III

    I tried substitution but I dont think it works
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  2. #2
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    Dear ceroseven,

    $\displaystyle 2sin^{2}x+sinx=1$

    $\displaystyle When, x=30^{o}\Rightarrow{L.H.S=2sin^{2}30^{o}+Sin30^{o} =\left(2\times{\frac{1}{4}}\right)+\frac{1}{2}=1}$

    $\displaystyle When, x=150^{o}\Rightarrow{L.H.S=2sin^{2}150^{o}+Sin150^ {o}=2sin^{2}30^{o}+sin30^{o}}$

    $\displaystyle Therefore, L.H.S=2sin^{2}150^{o}+Sin150^{o}=\left(2\times{\fr ac{1}{4}}\right)+\frac{1}{2}=1$

    $\displaystyle When, x=270^{o}\Rightarrow{L.H.S=2sin^{2}270^{o}+Sin270^ {o}=\left(2\times{-1}\right)+\left(-1\right)=-3}$

    Hope this helps.
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  3. #3
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    Quote Originally Posted by ceroseven View Post
    1. Which of the following is a solution to the equation 2sin^2x+sinx=1
    I. x=30
    II. x=150
    III. x=270

    a)III only
    b)I and III only
    c) II and III only
    d)I, II, III

    I tried substitution but I dont think it works
    $\displaystyle 2\sin^2{x} + \sin{x} = 1$

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


    Let $\displaystyle X = \sin{x}$ so that the equation becomes

    $\displaystyle 2X^2 + X - 1 = 0$

    $\displaystyle 2X^2 + 2X - X - 1 = 0$

    $\displaystyle 2X(X + 1) - 1(X + 1) = 0$

    $\displaystyle (X + 1)(2X - 1) = 0$

    $\displaystyle X + 1 = 0$ or $\displaystyle 2X - 1 = 0$

    $\displaystyle X = -1$ or $\displaystyle X = \frac{1}{2}$.


    So $\displaystyle \sin{x} = -1$ or $\displaystyle \sin{x} = \frac{1}{2}$

    $\displaystyle x = 270^\circ + 360^\circ n$ or $\displaystyle x = \left\{ 30^\circ, 150^\circ \right\} + 360^\circ n$, where $\displaystyle n$ represents the number of times you have gone around the unit circle.


    Can you see which solutions fit now?
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