<s>[INST] How do I solve the equation $2\sin(3x)\cos(4x)=1$? How do I solve the equation $2\sin(3x)\cos(4x)=1$? Normally, I'd use the $\sin(2x)=2\sin(x)\cos(x)$ rule, but here you have two different values for $x$, so I'm not sure how to add them together. 
 [/INST] Let $f(x)=2\sin(3x)\cos(4x)-1$ and solve $f(x)=0$.
The strategy will be to solve this using a Chebychev polynomial of the first type.
First, get a similar function entirely in terms of cosine by shifting the graph of $f$  horizontally to the right by $\frac{\pi}{2}$ units. This will be an even function with symmetric solutions. 
\begin{eqnarray}
g(x)&=&f\left(x-\frac{\pi}{2}\right)-1\\
&=&2\sin\left(3x-\frac{3\pi}{2}\right)\cos\left(4x-2\pi\right)-1\\
&=&2\cos(3x)\cos(4x)-1
\end{eqnarray}
We will find solutions for $g(x)=0$ then subtract $\frac{\pi}{2}$ units to obtain the solutions for $f(x)=0$.
Using the identities $2\cos A\cos B=\cos(A+B)+\cos(A-B)$ and $\cos(-\theta)=\cos(\theta)$, this becomes
$$ g(x)=\cos(7x)+\cos(x)-1 $$
If we let $u=\cos(x)$ then we can solve $g(x)=0$ by solving 
$$ T_7(u)+u-1=0 $$
where $T_7(u)$ is the seventh Chebychev polynomial of type 1. From Wolfram, 
$$ T_7(u)=64u^7-112u^5+56u^3-7u $$
so we need the solutions for
$$ 64u^7-112u^5+56u^3-6u-1=0 $$
Wolfram gives three real and four complex solutions. The real solutions are
$u=\frac{1}{2},\,0.759756,\,0.975695$.
So the solutions for $g(x)=0$ are
$$ \left\{2\pi n\pm\frac{\pi}{3},\,2\pi n\pm\arccos(0.759756),\,2\pi n\pm\arccos(0.975695) \right\} $$
To obtain the corresponding solutions for $f(x)=0$ we subtract $\frac{\pi}{2}$ from each of these six solutions.
$$ \left\{\left(2n-\frac{1}{2}\right)\pi\pm\frac{\pi}{3},\,\left(2n-\frac{1}{2}\right)\pi\pm\arccos(0.759756),\,\left(2n-\frac{1}{2}\right)\pi\pm\arccos(0.975695) \right\} $$
Letting $n=1$ gives the six solutions in the interval $[0,2\pi]$
$$\left\{\frac{11\pi}{6},\,\frac{7\pi}{6},\,5.4202475,\,4.0045304,\,4.9333148,\,4.4914632 \right\}  $$
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