How do I solve the equation $2\sin(3x)\cos(4x)=1$?

Question

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.

Answer 1

$$1=\sin(4x+3x)-\sin(4x-3x)=\sin7x-\sin x$$

$$\sin x+\dfrac12=\sin7x-\dfrac12$$

$$\sin x+\sin30^\circ=\sin7x+\sin210^\circ$$

$$2\sin\dfrac{x+30^\circ}2\cos\dfrac{x-30^\circ}2=2\sin\dfrac{7(x+30^\circ)}2\cos\dfrac{7(x-30^\circ)}2$$

$$\sin A\cos B=\sin7A\cos7B$$ where $2A=x+30^\circ,2B=x-30^\circ\implies A-B=30^\circ$

which is in general intractable

See Relationship among $A,B,C,D$ for $\cos A\cos B=\cos C\cos D$

Some of the trivial cases are supplied below:

Case $\#1:$

If $\sin A=\sin7A=0$

$\implies A=180^\circ m$

Else $\cos B=\cos7B=0$

$\implies B=(2n+1)90^\circ$

Case $\#2:$

$\sin A=0=\sin7A$ has been covered in case $\#1$

$\cos B=\cos7B=0$ has been covered in case $\#1$

$\sin A=\sin7A\iff\cos B=\cos7B$

$0=\sin7A-\sin A=2\sin3A\cos4A$

$0=\cos B-\cos7B=2\sin3B\sin4B$

Case $\#3:$

If $\sin A=-\sin7A$

$\sin A=0=\sin7A$ has been covered in case $\#1$

$\cos B=\cos7B=0$ has been covered in case $\#1$

Otherwise, $\sin A=-\sin7A\iff\cos B=-\cos7B$

$0=\sin7A+\sin A=2\sin4A\cos3A$

$0=\cos B+\cos7B=2\cos3B\cos4B$

Answer 2

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 $[0,2\pi]$

$$\left\{\frac{11\pi}{6},\,\frac{7\pi}{6},\,5.4202475,\,4.0045304,\,4.9333148,\,4.4914632 \right\} $$