I can simplify it all the way down to cos(6x) + sin(3x) = 0, however I get stuck hereCode:Find all solutions of 2 cos(3x)^2 + sin(3x)
I don't believe your expression is correct. Try it this way:
$\displaystyle sin^2(3x) + cos^2(3x) = 1$
So
$\displaystyle cos^2(3x) = 1 - sin^2(3x)$
So your equation becomes:
$\displaystyle 2 ( 1 - sin^2(3x) ) + sin(3x) = 0$
$\displaystyle 2sin^2(3x) - sin(3x) - 2 = 0$
which you can treat as a standard quadratic.
-Dan
Hello, Ccrow!
I'll take a guess at what you meant . . .
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} $