• Nov 21st 2006, 08:28 PM
Ccrow
Code:

Find all solutions of 2 cos(3x)^2 + sin(3x)
I can simplify it all the way down to cos(6x) + sin(3x) = 0, however I get stuck here :(
• Nov 22nd 2006, 03:28 AM
topsquark
Quote:

Originally Posted by Ccrow
Code:

Find all solutions of 2 cos(3x)^2 + sin(3x)
I can simplify it all the way down to cos(6x) + sin(3x) = 0, however I get stuck here :(

I don't believe your expression is correct. Try it this way:
$\displaystyle sin^2(3x) + cos^2(3x) = 1$

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

$\displaystyle 2 ( 1 - sin^2(3x) ) + sin(3x) = 0$

$\displaystyle 2sin^2(3x) - sin(3x) - 2 = 0$

which you can treat as a standard quadratic.

-Dan
• Nov 22nd 2006, 06:45 AM
Soroban
Hello, Ccrow!

I'll take a guess at what you meant . . .

Quote:

Find all solutions of: .$\displaystyle 2\cos^2(3x) + \sin(3x)\:=\:0$

Replace $\displaystyle \cos^2(3x)$ with $\displaystyle 1 - \sin^2(3x)$

. . then: .$\displaystyle 2\left[1 - \sin^2(3x)\right] + \sin(3x) \;=\;0$

And we have the quadratic: .$\displaystyle 2\sin^2(3x) - \sin(3x) - 2 \;=\;0$

Quadratic Formula: .$\displaystyle \sin(3x) \;= \;\frac{-(-1) \pm\sqrt{(-1)^2 - 4(2)(-2)}}{2(2)} \;= \;\frac{1 \pm\sqrt{17}}{4}$

We must discard the positive square root: .$\displaystyle \sin(3x) \,\neq\,1.28$

So we have: .$\displaystyle \sin(3x) = -0.7808\quad\Rightarrow\quad 3x = \begin{Bmatrix}-0.8959 + 2\pi n \\ 4.0375 + 2\pi n\end{Bmatrix}$

Therefore: .$\displaystyle x\:=\:\begin{Bmatrix}-0.2986 + \frac{2\pi}{3}n \\ 1.3458 + \frac{2\pi}{3}n\end{Bmatrix}$

• Nov 26th 2006, 03:41 PM
putnam120
Soroban you can have $\displaystyle \sin(3x)=1.28$ but you will be required to use complex numbers. I assume that he only wanted the real solutions any way, but just thought id point that out.