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$\sin(x)+\sin(3x)=0$

So to solve this I tried the following:

-First I transformed this expression from sum to product because it equals zero

$\sin(x)+\sin(3x)=0 => 2\sin(\frac{x+3x}{2})\cdot\cos(\frac{x-3x}{2})=0$

-Second part I got to

$2\sin(2x)\cos(x)=0$

-Third part: I have separated the equation into two parts:

$2\sin(2x)=0$ and $\cos(x)=0$

From now on I don't know how to get to the correct answers, which option should i circle and why:

a) $x=(2k+1)\frac{\pi}{2}$ or $x=\frac{k\pi}{2}$

b) $x=(k+1)\frac{\pi}{2}$ or $x=\frac{k\pi}{2}$

c) $x=(3k+1)\frac{\pi}{3}$ or $x=\frac{3k\pi}{2}$

d) $x=(k-1)\frac{\pi}{2}$ or $x=\frac{3k\pi}{2}$

where $k$ is any integer.

Bob Krueger
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L.B
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6 Answers6

3

Hint: The zeros of $\sin(x)$ occur at $k\pi$ for integers $k$. The zeros of $\cos(x)$ occur at $\frac{2k+1}{2}\pi$ for integers $k$.

Full solution: The zeros of $\cos(x) = 0$ occur where $x = \frac{2k+1}{2}\pi$ for some integer $k$. And the zeros of $2\sin(2x) = 0$ occur where $2x = k\pi$ for some integer $k$, i.e. where $x = \frac{k\pi}{2}$ for some integer $k$. Thus the answer is (a).

Bob Krueger
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aras
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2

\begin{align*}\sin (2x)=0 &\Longrightarrow 2x=k\pi\to x=\frac{k\pi}{2}\\ \cos x=0&\Longrightarrow x= \frac{\pi}{2}+k\pi=\frac{(2k+1)\pi}{2}\end{align*}

José Carlos Santos
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Arnaldo
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  • Could you please tell me how your answer is different from @expiTTpzo, it looks like he has 4 answers, I'm confused. – L.B Jul 03 '17 at 23:53
  • @L.B: He got the same two answers, but he wrote in a different way. $n\pi$ and $(2n+1)\pi/2$ give you the same set of solution as $k\pi/2$ and $(2k+1)\pi/2$. – Arnaldo Jul 04 '17 at 01:04
1

How I would have approached this problem,

$$\sin 3x = 3\sin x - 4\sin^3x$$

$$\sin x + \sin 3x = 4\sin x - 4 \sin^3x = 0 \implies \sin x(1-\sin^2 x) = 0 \implies \sin x \cos^2x = 0$$

$$\implies x = n\pi \text{ or } x = n\pi+\pi/2$$

$$\implies x = 2n\frac{\pi}{2} \text{ or } \frac{(2n+1)\pi}{2} \implies x = k\frac{\pi}{2} \text{ or } \frac{(k+1)\pi}{2}$$

Dhruv Kohli
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1

Hint:

$$ \sin \alpha =-\sin \beta \quad \iff \{\quad \alpha=-\beta+2n\pi \} \quad \mbox{or} \quad \{\alpha=\beta+(2n+1) \pi\} $$

with$, \quad n \in \mathbb{Z}$

Emilio Novati
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  • If $\sin\alpha = -\sin\beta = \sin(-\beta)$, then $$\alpha = -\beta + 2n\pi, n \in \mathbb{Z}$$ or $$\alpha = \pi - (-\beta) + 2n\pi, n \in \mathbb{Z} = \beta + (2n + 1)\pi, n \in \mathbb{Z}$$ – N. F. Taussig Jul 03 '17 at 19:50
  • Yes, thank you!! I forgot a set of solutions... I edit – Emilio Novati Jul 03 '17 at 20:03
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We need to solve $$\sin{x}=\sin(-3x),$$ which gives $x=-3x+2\pi k$ or $x=\pi-(-3x)+2\pi k$, where $k\in\mathbb Z$, which is $x=\frac{\pi k}{2}$ or $x=-\frac{\pi}{2}+\pi k$, which gives the answer: $$\left\{\frac{\pi k}{2}\big|k\in\mathbb Z\right\}$$

Michael Rozenberg
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0

Factorizing $2\sin(2x)\cos(x)$ I got $$4\cos(x)^2\sin(x)=0.$$ Therefore we get $\sin(x)=0$ or $\cos(x)=0$.

Bob Krueger
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Dr. Sonnhard Graubner
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