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Thread: Finding exact solutions in the interval

  1. #1
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    Question Finding exact solutions in the interval

    Hello,

    I am having trouble with this question, it say's:

    Find all exact solutions in the interval [0,2[pi]).

    a.) tan x= -2sin x

    b.) sin^2 θ + 5cosθ +1=0

    c.) sin^2 θ = -cos2θ

    Any help would be greatly appreciated =)
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  2. #2
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    Quote Originally Posted by KAY03 View Post
    Hello,

    I am having trouble with this question, it say's:

    Find all exact solutions in the interval [0,2[pi]).

    a.) tan x= -2sin x

    b.) sin^2 θ + 5cosθ +1=0

    c.) sin^2 θ = -cos2θ

    Any help would be greatly appreciated =)
    a) $\displaystyle \tan{x} = -2\sin{x}$

    $\displaystyle \frac{\sin{x}}{\cos{x}} = -2\sin{x}$

    $\displaystyle \frac{\sin{x}}{\sqrt{1 - \sin^2{x}}} = -2\sin{x}$

    $\displaystyle \frac{\sin^2{x}}{1 - \sin^2{x}} = 4\sin^2{x}$

    $\displaystyle \sin^2{x} = 4\sin^2{x}(1 - \sin^2{x})$

    $\displaystyle \sin^2{x} = 4\sin^2{x} - 4\sin^4{x}$

    $\displaystyle 0 = 3\sin^2{x} - 4\sin^4{x}$

    $\displaystyle 0 = \sin^2{x}(3 - 4\sin^2{x})$

    $\displaystyle \sin^2{x} = 0$ or $\displaystyle 3 - 4\sin^2{x} = 0$.


    Case 1:

    $\displaystyle \sin^2{x} = 0$

    $\displaystyle \sin{x} = 0$

    $\displaystyle x = \left\{0, \pi\right\} + 2\pi n$ where $\displaystyle n \in \mathbf{Z}$.


    Case 2:

    $\displaystyle 3 - 4\sin^2{x} = 0$

    $\displaystyle 4\sin^2{x} = 3$

    $\displaystyle \sin^2{x} = \frac{3}{4}$

    $\displaystyle \sin{x} = \pm \frac{\sqrt{3}}{2}$

    $\displaystyle x = \left\{\frac{\pi}{3}, \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, 2\pi - \frac{\pi}{3}\right\} + 2\pi n$ where $\displaystyle n \in \mathbf{Z}$

    $\displaystyle x = \left\{ \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\right\} + 2\pi n$ where $\displaystyle n \in \mathbf{Z}$.



    Putting it all together, we find

    $\displaystyle x = \left\{0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} \right\} + 2\pi n$ where $\displaystyle n \in \mathbf{Z}$.


    In the interval $\displaystyle x \in [0, 2\pi)$, the solutions are:

    $\displaystyle x = \left\{0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} \right\} + 2\pi n$ where $\displaystyle n \in \mathbf{Z}$.


    You may want to check these answers against the graph of $\displaystyle \tan{x} + 2\sin{x}$ though. Since I obtained the answers through squaring, it's possible that there are extraneous roots.
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    Quote Originally Posted by KAY03 View Post
    Hello,

    I am having trouble with this question, it say's:

    Find all exact solutions in the interval [0,2[pi]).

    a.) tan x= -2sin x

    b.) sin^2 θ + 5cosθ +1=0

    c.) sin^2 θ = -cos2θ

    Any help would be greatly appreciated =)
    (b) $\displaystyle \sin^2{\theta} + 5\cos{\theta} + 1 = 0$

    $\displaystyle 1 - \cos^2{\theta} + 5\cos{\theta} + 1 = 0$

    $\displaystyle -\cos^2{\theta} + 5\cos{\theta} + 2 = 0$

    $\displaystyle \cos^2{\theta} - 5\cos{\theta} - 2 = 0$.

    If we let $\displaystyle X = \cos^{\theta}$ we have a quadratic:

    $\displaystyle X^2 - 5X - 2 = 0$

    $\displaystyle X^2 - 5X + \left(-\frac{5}{2}\right)^2 - \left(-\frac{5}{2}\right)^2 - 2 = 0$

    $\displaystyle \left(X - \frac{5}{2}\right)^2 - \frac{25}{4} - \frac{8}{4} = 0$

    $\displaystyle \left(X - \frac{5}{2}\right)^2 - \frac{33}{4} = 0$

    $\displaystyle \left(X - \frac{5}{2}\right)^2 = \frac{33}{4}$

    $\displaystyle X - \frac{5}{2} = \pm \frac{\sqrt{33}}{2}$

    $\displaystyle X = \frac{5 \pm \sqrt{33}}{2}$.


    So $\displaystyle \cos{\theta} = \frac{5 + \sqrt{33}}{2}$ or $\displaystyle \cos{\theta} = \frac{5 - \sqrt{33}}{2}$.

    You will need to use technology to find $\displaystyle \theta$ now.
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    Quote Originally Posted by KAY03 View Post
    Hello,

    I am having trouble with this question, it say's:

    Find all exact solutions in the interval [0,2[pi]).

    a.) tan x= -2sin x

    b.) sin^2 θ + 5cosθ +1=0

    c.) sin^2 θ = -cos2θ

    Any help would be greatly appreciated =)
    (c) $\displaystyle \sin^2{\theta} = -\cos{2\theta}$

    $\displaystyle \sin^2{\theta} = -(\cos^2{\theta} - \sin^2{\theta})$

    $\displaystyle \sin^2{\theta} = \sin^2{\theta} - \cos^2{\theta}$

    $\displaystyle 0 = -\cos^2{\theta}$

    $\displaystyle 0 = \cos^2{\theta}$

    $\displaystyle 0 = \cos{\theta}$

    $\displaystyle \theta = \frac{\pi}{2} + \pi n$ where $\displaystyle n \in \mathbf{Z}$.


    So in the interval $\displaystyle \theta \in [0, 2\pi)$

    $\displaystyle \theta = \left\{\frac{\pi}{2}, \frac{3\pi}{2} \right\}$.
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