Finding the exact values of $\sin 4x - \sin 2x = 0$

Question

So I've used the double angle formula to turn

$$\sin 4x - \sin 2x = 0$$

$$2\sin2x\cos2x - \sin2x = 0$$

$$\sin2x(2\cos2x - 1) = 0$$

$$\sin2x = 0$$

$$2x = 0$$

$$x = 0$$

$$2\cos2x - 1 = 0$$

$$2\cos2x = 1$$

$$\cos2x = \frac{1}{2}$$

$$2x = 60$$

$$x = 30$$

With this information I am able to use the unit circle to find

$$x = 0,30,180,330,360$$

However, when I looked at the answers, $$x = 0,30,90,150,180,210,270,330,360$$

Can someone tell me how to obtain the rest of the values for $x$.

Thanks

One problem is that when $\sin(2x)=0$, there are more solutions than $2x=0$. In fact you may have $2x=0$, $2x=180$, $2x=360$, ... — Michael Burr, Dec 24 '15 at 13:12
@MichaelBurr I cant see any more values for sin(2x) = 0 other than 0,180 and 360.
But looking at what you just said, am i supposed to divide my current values by 2 to create 90?

2x = 180 x=90 ? — Kinson Chan, Dec 24 '15 at 13:27
Remember that sine is periodic, so you might also need $2x=540$ and $2x=720$ because, after dividing by $2$, $x$ is, at most, $360$. — Michael Burr, Dec 24 '15 at 13:34
@MichaelBurr How do i find the rest of the values for cos(2x) = 1/2 ? — Kinson Chan, Dec 24 '15 at 13:55
Basically, you must memorize important values of sine and cosine. — Michael Burr, Dec 24 '15 at 13:56
i know cosx = 30 and that it includes 30,150,210,330. But aren't 150 and 210 considered to be negative because they aren't in quad A or C in the unit circle? @MichaelBurr — Kinson Chan, Dec 24 '15 at 13:59

score 2 · Answer 1 · edited Dec 24 '15 at 13:39

2

Hint: $\cos(2x) = \frac{1}{2}$ has other solutions in the interval $[0, 360)$ besides $30$. What are they? Does $\sin(2x) = 0$ have any others besides what you find? Ask yourself what values of $2x$ you have to look at to find all of the values of $f(x)$ for $x \in [0, 360)$. Does this help?

An alternative (less formal) approach to this problem is to think about what $\sin(4x)$ and $\sin(2x)$ look like. From this it is clear that there will need to be $8$ solutions in $[0, 360)$ and a little thought will tell you how they have to be spaced.

edited Dec 24 '15 at 13:39

Kamil Jarosz

4,984

answered Dec 24 '15 at 13:15

Chris Dugan

186

I understand how there are other values for cosx = 30. But aren't the other values considered to be negative for it as they arent in the same quadrant in the unit circle? Right now I am only considering quadrant A and quadrant C for cosx. However, I can't see any other values for sin2x = 0 other than 0, 180 and 360 – Kinson Chan Dec 24 '15 at 13:32

Harish Chandra Rajpoot · Answer 2 · 2015-12-24T14:46:36.200

2

Notice, you should consider all the possible values of $x$ in the respective interval, $$\sin 2x(2\cos 2x-1)=0$$ consider the following two cases,

$$\sin 2x=0$$$$\implies 2x=n\pi\ \ \ or\ \ \ x=\frac{n\pi}{2}$$ Where, $n$ is any integer

Now, for the interval $x\in [0, 2\pi]$, setting $n=0, 1, 2, 3, 4$ one should get $$x=\color{blue}{0, 90^\circ, 180^\circ, 270^\circ, 360^\circ}$$

$$2\cos 2x-1\iff \cos 2x=\frac{1}{2}=\cos\frac{\pi}{3} $$ $$\implies 2x=2n\pi\pm\frac{\pi}{3}\ \ \ or\ \ \ x=n\pi\pm\frac{\pi}{6}$$

Now, for the interval $x\in [0, 2\pi]$, setting $n=0, 1, 2$ one should get $$x=\color{blue}{30^\circ, 150^\circ, 210^\circ, 330^\circ}$$

Hence, writing the complete solution for $x$, one should get
$$x=\color{red}{0, 30^\circ, 90^\circ, 150^\circ, 180^\circ, 210^\circ, 270^\circ, 330^\circ, 360^\circ}$$

edited Dec 24 '15 at 14:46

answered Dec 24 '15 at 13:32

Harish Chandra Rajpoot

37,450

I understand your cases, but how did you apply the nπ into the equation and give n certain numbers? @Harish-Chandra-Rajpoot – Kinson Chan Dec 24 '15 at 14:04
good question, sorry, I should have mentioned that $n$ is any integer hence select all the integer values of $n$ such that the values of $x$ obtained should lie in the respective interval as mentioned in the answer – Harish Chandra Rajpoot Dec 24 '15 at 14:09

Madrit Zhaku · Answer 3 · 2015-12-24T13:30:38.277

1

Use the formula: $$\sin x-\sin y=2\sin\frac{x-y}{2}\cos\frac{x+y}{2},$$ we have:

$$\sin 4x-\sin 2x=2\sin\frac{4x-2x}{2}\cos\frac{4x+2x}{2}=2\sin x\cos 3x.$$ Now, $$\sin 4x-\sin 2x=0$$ $$2\sin x\cos 3x=0$$ $$\sin x\cos 3x=0$$ $$\sin x=0\Rightarrow x=k\pi, k\in\mathbb{Z},$$ and $$\cos 3x=0\Rightarrow 3x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{3}, k\in\mathbb{Z}$$

edited Dec 24 '15 at 13:30

answered Dec 24 '15 at 13:20

Madrit Zhaku

5,294

score 0 · Answer 4 · answered Dec 24 '15 at 13:09

$$\sin(4x)-\sin(2x)=0\Longleftrightarrow$$ $$\sin(4x)=\sin(2x)$$

Take the inverse sine of both sides:

$$4x=\pi-2x+2\pi n_1\Longleftrightarrow$$ $$6x=\pi+2\pi n_1\Longleftrightarrow$$ $$x=\frac{\pi}{6}+\frac{\pi n_1}{3}$$
$$4x=2x+2\pi n_2\Longleftrightarrow$$ $$2x=2\pi n_2\Longleftrightarrow$$ $$x=\pi n_2$$

With $n_1,n_\in\mathbb{Z}$

So we got the following solutions:

$$x=\frac{\pi n}{2}$$ $$x=\pi n\pm\frac{\pi}{6}$$

score 0 · Answer 5 · answered Dec 24 '15 at 13:24

Using the prosthaphaeresis formula, $$ \sin{A}-\sin{B} = 2\cos{\frac{A+B}{2}}\sin{\frac{A-B}{2}}, $$ we find $$ 2\cos{3x}\sin{x}=0. $$ Now you just have to solve $\sin{x}=0$ or $\cos{3x}=0$. Sine has zeros at $180n$ degrees for $n$ an integer, and $\cos{y}$ has zeros at $y=180n+90$ degrees. Dividing by $3$ and checking all cases gives the result you want.

score 0 · Answer 6 · answered Dec 24 '15 at 13:36

First you should use the natural angle unit, which is the radian. Second, you should write the general solution, which is always a congruence.

Your equation factors as \begin{align*} 2\sin x\cos x(2\cos 2x -1)=0 &\iff\begin{cases}\sin x=0\\ \cos x=0 \\\cos x=\frac12 \end{cases}\\ & \iff \begin{cases} x\equiv 0\mod \pi \\ x\equiv \frac\pi2\mod\pi\\2x\equiv\pm\frac\pi3\mod 2\pi\iff x\pm\frac\pi6\mod \pi \end{cases} \end{align*}

Finding the exact values of $\sin 4x - \sin 2x = 0$

6 Answers6