# Solving 3 Different Trig Equations (find variable)

• Dec 17th 2006, 10:13 AM
SportfreundeKeaneKent
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:
http://img329.imageshack.us/img329/5...trigrw7.th.png

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

-Thanks
• Dec 17th 2006, 10:36 AM
CaptainBlack
Quote:

Originally Posted by SportfreundeKeaneKent
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:
http://img329.imageshack.us/img329/5...trigrw7.th.png

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
• Dec 17th 2006, 12:32 PM
Soroban
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.

Quote:

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

• Dec 18th 2006, 03:05 PM
Soroban
Hello again, SportfreundeKeaneKent!

Here's #3 . . .

Quote:

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

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