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Math Help - trig solving for x

  1. #1
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    trig solving for x

    1. solve for x in sinx tanx+ tanx- 2sinx+ cosx=0 for 0 <or equalx<orequal to 2\pi rads

    2. solve for x in \frac{1+sinx}{cosx}+\frac{cosx}{1+sinx}=4 for 0<or equal x <orequal 2\pi rads

    3. find the exact value of

    a) sin (\frac{\pi}{4}-\frac{\pi}{4})<br />

    b) sin (\frac{13\pi}{12})
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  2. #2
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    Hello, william!

    2. Solve for x\!:\;\;\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x}\:=\:4\qquad \text{for }0 \leq x \leq 2\pi

    Multipy by \cos x(1 + \sin x)\!:\quad (1+\sin x)^2 + \cos^2\!x \;=\;4\cos x(1 + \sin x)

    . . 1 + 2\sin x + \underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} \;=\;4\cos x + 4\sin x\cos x

    . . . . . . . . 2 + 2\sin x \;=\;4\cos x + 4\sin x\cos x

    Divide by 2: . 1 + \sin x \;=\;2\cos x + 2\sin x\cos x

    Re-arrange: . 2\sin x\cos x - \sin x + 2\cos x - 1 \;=\;0

    Factor: . \sin x(2\cos x - 1) + (2\cos x - 1) \;=\;0

    Factor: . (\sin x + 1)(2\cos x - 1) \:=\:0


    And we have:

    . . \sin x +1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:-1 \quad\Rightarrow\quad x \:=\:{\color{red}\rlap{//}}\frac{3\pi}{2} .
    extraneous root

    . . 2\cos x - 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:\frac{1}{2} \quad\Rightarrow\quad \boxed{x \:=\:\frac{\pi}{3},\:\frac{5\pi}{3}}




    3. Find the exact value of:

    a)\;\;\sin\left(\frac{\pi}{4}-\frac{\pi}{4}\right)
    \sin\left(\frac{\pi}{4} - \frac{\pi}{4}\right) \;=\;\sin(0) \;=\;0


    b)\;\;\sin \left(\frac{13\pi}{12}\right)
    Note that: . \frac{13\pi}{12} \;=\;\frac{9\pi}{12} + \frac{4\pi}{12} \;=\;\frac{3\pi}{4} + \frac{\pi}{3}

    We have: . \sin\left(\frac{13\pi}{12}\right)  \;=\;\sin\left(\frac{3\pi}{4} + \frac{\pi}{3}\right)

    . . . . . . = \;\;\sin\left(\frac{3\pi}{4}\right)\cos\left(\frac  {\pi}{3}\right) + \cos\left(\frac{3\pi}{4}\right)\sin\left(\frac{\pi  }{3}\right)

    . . . . . . = \;\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}  \right) + \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\  right) \;\;=\;\;\frac{1-\sqrt{3}}{2\sqrt{2}} \;\;=\;\;\frac{\sqrt{2} - \sqrt{6}}{4}

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