# Thread: Trigonometry monster equations! Help

1. ## Trigonometry monster equations! Help

Hello Everyone,

thanks already for reading, and if you can help me out, even more thanks!

My teacher surprised us today with extra credit homework. He said we can ask anyone for help. He thinks that not too many people will be able to solve his problems.

I will list them all and number them. If any one of these equations look like you can do it, feel free to post the answer and the corresponding number, so that questions won't be answered twice.

Thanks so much!

1) $\displaystyle cos7x * cos13x = cosx * cos19x$
2)$\displaystyle sin2x * cos8x + sin6x = 0$
3)$\displaystyle cos9x - cos7x + cos3x - cosx = 0$
4)$\displaystyle 2sin^2(x) - cos2x + sin2x = 0$
5)$\displaystyle 2cosx + 3sinx = 1$
7)$\displaystyle sinx + sin^2(x) + cos^3(x) = 0$

2. Hello, rexexdesign!

These are very tricky . . . Here's #6 . . .

$\displaystyle 6)\;\;4\sin^4\!x -1 \:= \:5\cos^2\!x$
We have: .$\displaystyle 4\sin^4\!x - 1 \:=\:5(1-\sin^2\!x)$

. . which simplifies to: .$\displaystyle 4\sin^4x + 5\sin^2\!x - 6 \;=\;0$

. . which factors: .$\displaystyle (\sin^2\!x + 2)(4\sin^2\!x - 3) \;=\;0$

Then:. $\displaystyle \sin^2\!x + 2 \:=\:0\quad\Rightarrow\quad \sin^2\!x \:=\:-2$ . . . which has no real roots

And: .$\displaystyle 4\sin^2\!x-3\:=\:0\quad\Rightarrow\quad \sin^2\!x\:=\:\frac{3}{4} \quad\Rightarrow\quad \sin x \:=\:\pm\frac{\sqrt{3}}{2}$

. . Therefore: .$\displaystyle x \;=\;\frac{\pi}{3}+ \pi n,\;\frac{2\pi}{3} + \pi n$

3. Soroban, that was a good eye. We have done something like this before in class, only much simpler.

For #1, how do I break this into smaller pieces: $\displaystyle cos7x * cos13x = cosx * cos19x$

do I have to use the double angle formula a whole bunch of times? Or is there another way to get the multiple angles split?

4. Hello, rexexdesign!

Your teacher is right . . . These are very messy problems.
They require familiarity with a variety of identities . . . and stamina.

$\displaystyle 1)\;\;\cos7x\cos13x \;= \;\cos x\cos19x$
I would use a product-to-sum identity: .$\displaystyle \cos A\cos B \;=\;2\cos\left(\frac{A+B}{2}\right)\cos\left(\fra c{A-B}{2}\right)$

We have: .$\displaystyle \cos13x\cos7x \:=\:\cos19x\cos x$

. . which becomes: .$\displaystyle 2\cos(10x)\cos(3x) \;=\;2\cos(10x)\cos(9x)$

. . and we have: .$\displaystyle \cos(10x)\cos(3x) - \cos(10x)\cos(9x) \;=\;0$

. . Factor: .$\displaystyle \cos(10x)\cdot\left[\cos(3x) - \cos(9x)\right] \;=\;0$

And we have two equations to solve:

$\displaystyle [1]\;\;\cos(10x) \:=\:0\quad\Rightarrow\quad 10x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2},\:\frac{5\pi}{ 2},\:\hdots \quad\Rightarrow\quad\boxed{x \;=\;\frac{\pi}{20},\:\frac{3\pi}{20},\:\frac{5\pi }{20},\:\hdots}$

$\displaystyle [2]\;\;\cos(3x) \:=\:\cos(9x)$

We will need a multiple-angle identitty: .$\displaystyle \cos3A \;=\;4\cos^3\!A - 3\cos A$

. . So that: .$\displaystyle \cos(9x) \;=\;4\cos^3\!(3x) - 3\cos(3x)$

The equation becomes: .$\displaystyle \cos(3x) \;=\;4\cos^3\!(3x) - 3\cos(3x)$

We have: .$\displaystyle 4\cos^3(3x) - 4\cos(3x) \:=\:0\quad\Rightarrow\quad 4\cos(3x)\cdot\left[\cos^2(3x) - 1\right] \:=\:0$

And we have two more equations to solve:

$\displaystyle [1]\;\;4\cos(3x) \:=\:0\quad\Rightarrow\quad 3x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2},\:\frac{5\pi}{ 2},\:\hdots \quad\Rightarrow\quad\boxed{ x \;=\;\frac{\pi}{6},\:\frac{\pi}{2},\:\frac{5\pi}{6 },\:\hdots}$

$\displaystyle [2]\;\;\cos^2(3x) \:=\:1\quad\Rightarrow\quad \cos3x \:=\:\pm1\quad\Rightarrow\quad 3x \;=\;0,\:\pi,\:2\pi,\:3\pi,\:4\pi,\:\hdots$

. . . $\displaystyle \boxed{x \;=\;0,\:\frac{\pi}{3},\:\frac{2\pi}{3},\:\pi,\:\f rac{4\pi}{3},\:\hdots}$

It is always good to check the answers.
Often, some are extraneous roots.
. . (No, I didn't check these . . .)

5. funny, I used the product to sum formula as well. For me the equation looked harder after I was done. I did not know that you can cancel $\displaystyle cos6x/2$ to $\displaystyle cos3x$, thus I ended up with an equation divided by two.
My above conclusion was derived from your solution, is that correct?

Thanks so much. I will check the roots to see if they work. I have until tomorrow afternoon to get these problems solved.

6. Originally Posted by Soroban

$\displaystyle [1]\;\;4\cos(3x) \:=\:0\quad\Rightarrow\quad 3x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2},\:\frac{5\pi}{ 2},\:\hdots \quad\Rightarrow\quad\boxed{ x \;=\;\frac{\pi}{6},\:\frac{\pi}{2},\:\frac{5\pi}{6 },\:\hdots}$

$\displaystyle [2]\;\;\cos^2(3x) \:=\:1\quad\Rightarrow\quad \cos3x \:=\:\pm1\quad\Rightarrow\quad 3x \;=\;0,\:\pi,\:2\pi,\:3\pi,\:4\pi,\:\hdots$

. . . $\displaystyle \boxed{x \;=\;0,\:\frac{\pi}{3},\:\frac{2\pi}{3},\:\pi,\:\f rac{4\pi}{3},\:\hdots}$

It is always good to check the answers.
Often, some are extraneous roots.
. . (No, I didn't check these . . .)

I checked them and they seemed OK to me