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Thread: 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:

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

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

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

    $\displaystyle (1-s^2)-3s^2=7s-1$,

    simplifying:

    $\displaystyle 4s^2+7s-2=0$.

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

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

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

    $\displaystyle x\approx 0.3$ and $\displaystyle x \approx 2.9$

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

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

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

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

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

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


    We have two equations to solve:

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

    . . $\displaystyle 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: .$\displaystyle \boxed{W \:\approx\:1.23,\:5.05}$


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

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

    However, the other answers do check out.

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

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  4. #4
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    Hello again, SportfreundeKeaneKent!

    Here's #3 . . .


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

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

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

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


    We have two equations to solve:

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

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

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