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Math Help - Solving 3 Different Trig Equations (find variable)

  1. #1
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    Solving 3 Different Trig Equations (find variable)

    Here are a few trig equations, all three of them involve finding the variable. I've gotten some different answers for all of them so I was wondering if anyone could help.

    By the way, I'm not sure on how to post images from image shack so could someone PM me on how to post a jpeg which has been uploaded to imageshack on this site.

    Here's the first attempt:


    If that doesn't work, here's the link to the three questions:
    http://img329.imageshack.us/img329/5343/z1trigrw7.png

    -Thanks
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  2. #2
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Here are a few trig equations, all three of them involve finding the variable. I've gotten some different answers for all of them so I was wondering if anyone could help.

    By the way, I'm not sure on how to post images from image shack so could someone PM me on how to post a jpeg which has been uploaded to imageshack on this site.

    Here's the first attempt:


    If that doesn't work, here's the link to the three questions:
    http://img329.imageshack.us/img329/5343/z1trigrw7.png

    -Thanks
    The first question is:

    \cos^2(x)-3\sin^2(x)=7\sin(x)-1

    (from now on I will use c, s for \cos(x) and \sin(x))

    Replace \cos^2(x) by 1-\sin^2(x) to get:

    (1-s^2)-3s^2=7s-1,

    simplifying:

    4s^2+7s-2=0.

    This may be solved using the quadratic formula to get: s=1/4, -2. Now the -2 root is not possible for real angles so we are left with \sin(x)=1/4.

    There are two solutions for this in the range [0,2\pi], they are:

    x=\arcsin(1/4) and x=\pi-\arcsin(1/4), or:

    x\approx 0.3 and x \approx 2.9

    RonL
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  3. #3
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    Hello, SportfreundeKeaneKent!

    All of these are of the same type: make a substitution and solve the quadratic.
    . . But I admit that the answers are tricky to check.


    2) Solve: . 1 + 2\cos W \;=\;\sqrt{3 + 2\cos W - \sin^2W} . . . given 0 \leq W \leq 2\pi

    Replace \sin^2W with 1 - \cos^2W

    . . 1 + 2\cos W \;= \;\sqrt{3 + 2\cos W - (1 - \cos^2W)} \;=\;\sqrt{\cos^2W + 2\cos W + 2}

    Square both sides: . 1 + 4\cos W + 4\cos^2W \;= \;\cos^2W + 2\cos W + 2

    . . and we have: . 3\cos^2W + 2\cos W - 1 \:=\:0

    . . which factors: . (\cos W + 1)(3\cos W - 1) \:=\:0


    We have two equations to solve:

    . . \cos W +1 \:=\:0\quad\Rightarrow\quad \cos W \:=\:-1\quad\Rightarrow\quad \boxed{W \,=\,\pi}

    . . 3\cos W - 1 \:=\:0\quad\Rightarrow\quad\cos W = \frac{1}{3}\quad\Rightarrow\quad W = \cos^{-1}\left(\frac{1}{3}\right)
    . . . . Hence: . \boxed{W \:\approx\:1.23,\:5.05}


    We must check these answers; squaring can produce extraneous roots.

    We find that W = \pi is not a solution.

    However, the other answers do check out.

    . . Solution: . \boxed{\boxed{W \:=\:1.23,\;5.05}}

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  4. #4
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    Hello again, SportfreundeKeaneKent!

    Here's #3 . . .


    3) Solve: . 8\sin^2x - 2\cos x \:=\:5 . for  0 \leq x \leq \frac{\pi}{2} . x is in Quadrant 1

    Replace \sin^2x with 1 - \cos^2x\!:\;\;8(1 - \cos^2x) - 2\cos x \:=\:5

    . . which simplifies to: . 8\cos^2x + 2\cos x - 3 \:=\:0

    . . which factors: . (4\cos x + 3)(2\cos x - 1) \:=\:0


    We have two equations to solve:

    . . 4\cos x + 3 \:=\:0\quad\Rightarrow\quad \cos x = -\frac{3}{4} . . . . then x is not in Quadrant 1

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

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