Is this the right approach to solve $\sin 4 \omega - \sin 6 \omega=0$ analytically?

Question

The problem is as follows:

$$\sin 4 \omega - \sin 6 \omega=0$$

What I did to solve this problem was to do a bit of algebraic work as shown below

$$\sin 4 \omega - \sin 6 \omega=0$$

$$2 \sin 2 \omega \cos 2 \omega - \left ( \sin 4 \omega \cos 2 \omega + \cos 4 \omega \sin 2 \omega \right )= 0$$

$$2 \sin 2 \omega \cos 2 \omega - \sin 4 \omega \cos 2 \omega - \cos 4 \omega \sin 2 \omega = 0$$

$$2 \sin 2 \omega \cos 2 \omega - 2 \sin 2 \omega \cos 2\omega \times \cos 2 \omega - \left ( \cos ^{2} 2 \omega - \sin ^{2} 2 \omega \right ) \sin 2 \omega = 0$$

$$2 \sin 2 \omega \cos 2 \omega - 2 \sin 2 \omega \cos 2\omega \times \cos 2 \omega - \left ( \cos ^{2} 2 \omega - \left (1-\cos ^{2} 2 \omega \right ) \right ) \sin 2 \omega = 0$$

$$2 \sin 2 \omega \cos 2 \omega - 2 \sin 2 \omega \cos ^{2} 2\omega - \left ( \cos ^{2} 2 \omega - 1+\cos ^{2} 2 \omega \right ) \sin 2 \omega = 0$$

$$2 \sin 2 \omega \cos 2 \omega - 2 \sin 2 \omega \cos ^{2} 2\omega - \left ( 2 \cos ^{2} 2 \omega - 1 \right ) \sin 2 \omega = 0$$

$$\sin 2 \omega \left ( 2 \cos 2 \omega - 2 \cos ^{2} 2 \omega - 2 \cos^{2} 2 \omega +1 \right ) = 0$$

$$\sin 2 \omega \left (- 4 \cos ^{2} 2 \omega + 2 \cos 2 \omega +1 \right ) = 0$$

$$- \sin 2 \omega \left ( 4 \cos ^{2} 2 \omega - 2 \cos 2 \omega - 1 \right ) = 0$$

Then dividing both sides by $ -1 $:

$$\sin 2 \omega \left ( 4 \cos ^{2} 2 \omega - 2 \cos 2 \omega - 1 \right ) = 0$$

Now here is where I'm in doubt:

Should I go on this route?

$\sin 2 \omega = 0$

Therefore: (in the interval of $[0, 2\pi ]$)

$2 \omega = 0$

$\omega = 0$

$2 \omega = \pi$

$\omega = \frac{\pi}{2}$

$2 \omega = 2 \pi$

$\omega = \pi$

$\omega = \left \{0, \frac{\pi}{2}, \pi \right \}$

Then in the other factor:

$$4 \cos ^{2} 2 \omega - 2 \cos 2 \omega - 1 = 0$$

$$\cos 2 \omega = \frac{-\left( -2 \right ) \pm \sqrt {(-2)^{2}-4(4)(-1)}}{2\times 4}$$

$$\cos 2 \omega = \frac{2 \pm \sqrt {20}}{8}=\frac{1 \pm \sqrt {5}}{4}$$

$$\cos 2 \omega = \frac{1 + \sqrt {5}}{4}$$

$$2 \omega = \cos^{-1} \left ( \frac{1 + \sqrt {5}}{4} \right )$$

$$\omega = \frac{\cos^{-1} \left ( \frac{1 + \sqrt {5}}{4} \right )}{2}$$

Since this value is positive then the other choice in the interval $[0, 2\pi ]$ would be:

$$\omega = 2 \pi - \frac{\cos^{-1} \left ( \frac{1 + \sqrt {5}}{4} \right )}{2} = \frac{4 \pi - \cos^{-1} \left ( \frac{1 + \sqrt {5}}{4} \right )}{2}$$

Finally for the second solution of the quadratic equation:

$$\cos 2 \omega = \frac{1 - \sqrt {5}}{4}$$

Since this value is negative, the angle $\omega$ must be in the second quadrant and the third quadrant, therefore:

$$\omega = \frac{\cos^{-1} \left ( \frac{1 - \sqrt {5}}{4} \right )}{2}$$

then the other solution in the interval $[0, 2\pi ]$ would be:

$$\omega = 2 \pi - \frac{\cos^{-1} \left ( \frac{1 - \sqrt {5}}{4} \right )}{2} = \frac{4 \pi - \cos^{-1} \left ( \frac{1 - \sqrt {5}}{4} \right )}{2}$$

Therefore the solution so far for these four would be:

$$\omega = \left \{ \frac{\cos^{-1} \left ( \frac{1 + \sqrt {5}}{4} \right )}{2}, \frac{4 \pi - \cos^{-1} \left ( \frac{1 + \sqrt {5}}{4} \right )}{2}, \frac{\cos^{-1} \left ( \frac{1 - \sqrt {5}}{4} \right )}{2} , \frac{4 \pi - \cos^{-1} \left ( \frac{1 - \sqrt {5}}{4} \right )}{2} \right \}$$

Then adding the other two which I obtained before would make $\omega$ this set:

$$\omega = \left \{ 0, \frac{\pi}{2}, \pi ,\frac{\cos^{-1} \left ( \frac{1 + \sqrt {5}}{4} \right )}{2}, \frac{4 \pi - \cos^{-1} \left ( \frac{1 + \sqrt {5}}{4} \right )}{2}, \frac{\cos^{-1} \left ( \frac{1 - \sqrt {5}}{4} \right )}{2} , \frac{4 \pi - \cos^{-1} \left ( \frac{1 - \sqrt {5}}{4} \right )}{2} \right \}$$

But my question arises whether if the preceding set is the correct way to go:

I had doubts whether should I use

$$\sin 2 \omega = 0$$

$$\sin 2 \omega = \sqrt{1 - \cos 2 \omega} = 0$$

$$\sqrt{1 - \cos 2 \omega} = 0$$

$$1 - \cos 2 \omega = 0$$

$$\cos 2 \omega = 1$$

$$\omega = \frac{\cos^{-1} \left ( 1 \right )}{2}= 0 , \frac {2 \pi}{2}$$

$$\omega = 0, \pi$$

But here the "answer" $\frac{\pi}{2}$ seems missing. Did I miss something?. I'd hope somebody can help me whether to look if my procedure was appropriate for this case. The reason why I decided to change the sine function into cosine for solving the equation was because I was not sure if is allowed to equate to zero both factors if the functions are different or do they have to be the same?. But changing from one function to another I seem to miss one answer. Is my logic in the right path?

Answer 1

We analyze the errors in your approach:

Your procedure was fine until you attempted to solve the equation $\sin 2\omega = 0$.

In your first approach, you missed the solutions $\omega = \dfrac{3\pi}{2}, 2\pi$.

In your second approach, \begin{align*} \sin^2 2\omega + \cos^2 2\omega & = 1\\ \sin^2 2\omega & = 1 - \cos^2 2\omega\\ |\sin 2\omega| & = \sqrt{1 - \cos^{\color{red}{2}} 2\omega}\\ \sin 2\omega & = \color{red}{\pm} \sqrt{1 - \cos^{\color{red}{2}} 2\omega}\\ 0 & = \color{red}{\pm} \sqrt{1 - \cos^{\color{red}{2}} 2\omega}\\ 0 & = 1 - \cos^{\color{red}{2}} 2\omega\\ \cos^{\color{red}{2}} 2\omega & = 1\\ |\cos 2\omega| & = 1\\ \cos 2\omega & = \color{red}{\pm} 1 \end{align*} which again yields the solutions $$\omega = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$$ The value $\omega = \dfrac{\pi}{2}$ clearly is an answer to the original equation since $$\sin 4\omega - \sin 6\omega = \sin\left(4 \cdot \frac{\pi}{2}\right) - \sin\left(6 \cdot \frac{\pi}{2}\right) = \sin(2\pi) - \sin(3\pi) = 0 - 0 = 0$$ Similarly, you can verify that $0$, $\pi$, $\dfrac{3\pi}{2}$, and $2\pi$ are solutions to the original equation by direct substitution.

In solving the equation $$4\cos^2 2\omega - 2\cos 2\omega + 1 = 0$$ you overlooked the fact that if $\omega \in [0, 2\pi]$, then $2\omega \in [0, 4\pi]$, so you missed four more solutions.

Since $\cos(-x) = \cos x$, $\cos 2\omega = \dfrac{1 + \sqrt{5}}{4} \implies$ $$2\omega = \arccos\left(\frac{1 + \sqrt{5}}{4}\right) + 2m\pi, m \in \mathbb{Z}$$ or $$2\omega = -\arccos\left(\frac{1 + \sqrt{5}}{4}\right) + 2n\pi, n \in \mathbb{Z}$$

Since $\omega \in [0, 2\pi]$, we require that $2\omega \in [0, 4\pi] \implies m \in \{0, 1\}$, which yields the solutions \begin{align*} 2\omega & = \arccos\left(\frac{1 + \sqrt{5}}{4}\right) & 2\omega & = \arccos\left(\frac{1 + \sqrt{5}}{4}\right) + 2\pi\\ \omega & = \frac{1}{2}\arccos\left(\frac{1 + \sqrt{5}}{4}\right) & \omega & = \frac{1}{2}\arccos\left(\frac{1 + \sqrt{5}}{4}\right) + \pi\\ \end{align*} and $n \in \{1, 2\}$, which yields the solutions \begin{align*} 2\omega & = -\arccos\left(\frac{1 + \sqrt{5}}{4}\right) + 2\pi & 2\omega & = -\arccos\left(\frac{1 + \sqrt{5}}{4}\right) + 4\pi\\ \omega & = -\frac{1}{2}\arccos\left(\frac{1 + \sqrt{5}}{4}\right) + \pi & \omega & = -\frac{1}{2}\arccos\left(\frac{1 + \sqrt{5}}{4}\right) + 2\pi \end{align*}

Similarly, $\cos 2\omega = \dfrac{1 - \sqrt{5}}{4} \implies$ $$2\omega = \arccos\left(\frac{1 - \sqrt{5}}{4}\right) + 2m\pi, m \in \mathbb{Z}$$ or $$2\omega = -\arccos\left(\frac{1 - \sqrt{5}}{4}\right) + 2n\pi, n \in \mathbb{Z}$$ Since $\omega in [0, 2\pi] \implies 2\omega \in [0, 4\pi]$, $m \in \{0, 1\}$, which yields the solutions \begin{align*} 2\omega & = \arccos\left(\frac{1 - \sqrt{5}}{4}\right) & 2\omega & = \arccos\left(\frac{1 - \sqrt{5}}{4}\right) + 2\pi\\ \omega & = \frac{1}{2}\arccos\left(\frac{1 - \sqrt{5}}{4}\right) & \omega & = \frac{1}{2}\arccos\left(\frac{1 - \sqrt{5}}{4}\right) + \pi\\ \end{align*} and $n \in \{1, 2\}$, which yields the solutions \begin{align*} 2\omega & = -\arccos\left(\frac{1 - \sqrt{5}}{4}\right) + 2\pi & 2\omega & = -\arccos\left(\frac{1 - \sqrt{5}}{4}\right) + 4\pi\\ \omega & = -\frac{1}{2}\arccos\left(\frac{1 - \sqrt{5}}{4}\right) + \pi & \omega & = -\frac{1}{2}\arccos\left(\frac{1 - \sqrt{5}}{4}\right) + 2\pi \end{align*}

An alternate solution:

Consider the following diagram.

Two directed angles in standard position (vertex at the origin, initial side on the positive $x$-axis) have the same sine if the $y$-coordinates of the points where the terminal sides of the angles intersect the unit circle are equal. By symmetry, $\sin\theta = \sin\varphi$ if $\varphi = \theta$ or $\varphi = \pi - \theta$. Since any angle coterminal with these angles has the same $y$-coordinate, $\sin\theta = \sin\varphi$ if $$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$ or $$\varphi = \pi - \theta + 2k\pi, k \in \mathbb{Z}$$

Let's return to the original problem. \begin{align*} \sin 4\omega - \sin 6\omega & = 0\\ \sin 4\omega & = \sin 6\omega \end{align*}

Setting $\theta = 4\omega$ and $\varphi = 6\omega$ yields \begin{align*} 6\omega & = 4\omega + 2m\pi, m \in \mathbb{Z} & 6\omega & = \pi - 4\omega + 2n\pi, n \in \mathbb{Z}\\ 2\omega & = 2m\pi, m \in \mathbb{Z} & 10\omega & = \pi + 2n\pi, n \in \mathbb{Z}\\ \omega & = m\pi, m \in \mathbb{Z} & \omega & = \frac{\pi}{10} + \frac{n\pi}{5}, n \in \mathbb{Z} \end{align*} Restricting our attention to the interval $[0, 2\pi]$ implies $m \in \{0, 1, 2\}$, which yields the solutions $$\omega = 0, \pi, 2\pi$$ and $n \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, which yields the solutions $$\omega = \frac{\pi}{10}, \frac{3\pi}{10}, \frac{\pi}{2}, \frac{7\pi}{10}, \frac{9\pi}{10}, \frac{11\pi}{10}, \frac{13\pi}{10}, \frac{3\pi}{2}, \frac{17\pi}{10}, \frac{19\pi}{10}$$ as you can verify by direct substitution and the use of symmetry.

Notice that with this method, you do not need to know that $$\arccos\left(\frac{1 + \sqrt{5}}{4}\right) = \frac{\pi}{5}$$ or that $$\arccos\left(\frac{1 - \sqrt{5}}{4}\right) = \frac{3\pi}{5}$$

Answer 2

You have this general theorem:

Theorem. Let $A,B$ be real numbers. Then $$AB = 0 \quad \iff \quad (A=0) \lor (B=0) $$

In plain words, it is necessary and sufficient that either $A = 0$ or $B = 0$ (or both) for $AB$ to also equal $0$. This means that your problem is logically equivalent to determining under which conditions $A = 0$ and under which (generally different) conditions $B =0$, and then taking the union of those conditions. Why the union? Let's say that $A$ and $B$ depend on $\omega$: only some of the $\omega$'s are such that $A(\omega) = 0$, and similarly only some of the $\omega$'s are such that $B(\omega) = 0$, and there may even be some $\omega$'s for which both $A(\omega) = B(\omega) = 0$. We want to take all of these $\omega$'s (we wouldn't want to leave any solutions behind), so that's why we take the union.

In your specific case, $A(\omega) = \cos(5\omega)$ and $B(\omega) = \sin\omega$, so that $$5\omega = \arccos0=\frac \pi 2 + k\pi,\quad \forall k \in \mathbb Z, \qquad \omega = \arcsin 0 = k\pi,\quad \forall k \in \mathbb Z $$ are the two conditions you're looking for. Ultimately, your equation is satisfied whenever $\omega = k \pi$ or $\omega = \frac \pi 5\left(\frac 1 2 + k\right)$, for some $k \in \mathbb Z$, that is, it is necessary and sufficient for $\omega$ to take either one of these forms for $\omega$ to be an acceptable solution to your equation. This means you have found all possible solutions to the original equation.

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A side note: at some point in your calculations, you were to calculate $$\arccos\left(\frac{1 + 2\sqrt 5}{4}\right) $$ but this is nonsensical, because the argument of $\arccos$ is a real number greater than $1$, and we know that the range of the cosine function is $[-1,1]$. You just have to ditch this case. If you switch the $+$ with a $-$, however, you do get a number lying in the range of the cosine, so it's all fine there – even though, if you plug it in your original equation, you do not get zero, so somewhere in your calculations you must have made a mistake...

Answer 3

A much simpler way: use the relation

$$\sin m \omega = \sin n \omega \implies n \omega = m \omega + 2 \pi k \text { or } n \omega = \pi - m \omega + 2 \pi k$$

If $m = 4$ and $n = 6$, you should have as solutions

$$\omega = (k - 1)\pi \vee \omega = \dfrac {(2k - 1) \pi}{10}, k \in \mathbb{Z}.$$

Answer 4

Another approach: $$ \begin{align} \sin(4\omega)-\sin(6\omega) &=\sin(5\omega-\omega)-\sin(5\omega+\omega)\\ &=-2\cos(5\omega)\sin(\omega) \end{align} $$ so we need either $$ \cos(5\omega)=0\iff5\omega=\left(k+\frac12\right)\pi $$ or $$ \sin(\omega)=0\iff\omega=k\pi $$ Therefore, we get the solutions $$ \omega\in\left\{\frac{2k+1}{10}\pi,k\pi:k\in\mathbb{Z}\right\} $$