Find the number of real solutions of equation : $\sin (3\theta) = 4\sin\theta\sin(2\theta) \sin(4\theta)$ in $\theta \in (0, \pi)$.

Question

My try : Sorry , but I can't figure out what is the first step to do in this question . But still I tried and converted $\sin(3\theta)$ into $\sin (2\theta + \theta)$ but failed to proceed . Please tell me it's approach.

Answer 1

There are $8$ solutions, namely $$0,\quad{\pi\over9},\quad{2\pi\over9},\quad{4\pi\over9},\quad{5\pi\over9},\quad{7\pi\over9},\quad{8\pi\over9},\quad\pi\ .\tag{1}$$ In order to arrive at them we write $e^{it}=:z$. The equation $$\bigl(f(t):=\bigr)\quad 4\sin t\sin(2t)\sin(4t)-\sin(3t)=0$$ then appears as $$z^{14}-z^{12}+z^8-z^6+z^2-1=0\ ,\tag{2}$$ or $$(z^2-1)(z^{12}+z^6+1)=0\ .$$ The first factor gives $z\in\{1,-1\}$. For the second factor we write $z^6=:w$ and then have to solve $w^2+w+1=0$. It follows that $w\in\{e^{2\pi i/3}, \>e^{-2\pi i/3}\}$. For each of these $w$-values we obtain $6$ possible values of $z$ forming a regular hexagon on the unit circle. In all we obtain $14$ different values $z_k$ satisfying $(2)$. Collecting the $z_k$ with argument $t_k\in\bigl[0,{\pi\over2}\bigr]$ leads to the list $(1)$. Here is a plot of the function $f$. In fact $$f(t)=-\sin t\bigl(2\cos(6t)+1\bigr)\ .$$

Answer 2

Use that

$$\sin(3x)=3\cos(x)^2\sin(x)-\sin(x)^3$$ and $$\sin(4x)=4\cos(x)^3\sin(x)-4\cos(x)\sin(x)^3$$

Answer 3

Let $\cos2\theta=x$.

Thus, we need to solve $$3-4\sin^2\theta=8\sin^22\theta\cos2\theta$$ or $$3-2(1-x)=8(1-x^2)x$$ or $$8x^3-6x+1=0$$ or $$2\cos6\theta+1=0$$ or $$\cos6\theta=-\frac{1}{2}.$$ Can you end it now?

On $(0,\pi)$ I got five roots: $$\left\{20^{\circ},40^{\circ},100^{\circ},140^{\circ},160^{\circ}\right\}$$

Answer 4

Let $u = \sin \theta$.

$$\begin{array}{rcl} \sin(3\theta) &=& 4 \sin (\theta) \sin (2\theta) \sin (4\theta) \\ 3u - 4u^3 &=& 4 u (2u\cos\theta) (4u\cos\theta - 8u^3\cos\theta) \\ 3u - 4u^3 &=& 4 u (2u) (4u - 8u^3) (1 - u^2) \\ 3u - 4u^3 &=& 64 u^7 - 96 u^5 + 32 u^3 \\ 64 u^7 - 96 u^5 + 36 u^3 - 3u &=& 0 \\ \end{array}$$

Since $u \ne 0$, let $t = u^2$.

$$64t^3 - 96t^2 + 36t - 3 = 0$$

We depress the cubic by letting $t := x + \frac12$:

$$64 x^3 - 12 x - 1 = 0$$

We consider $4(ax)^3 - 3(ax) = (4a^3) x^3 - (3a) x$ and want $a$ such that $4a^3 : -3a = 64 : -12$, i.e. $4a^2 : 3 = 16 : 3$, so we can let $a = 2$.

That means we further let $s := 2x$ and this gives us:

$$4s^3-3s = \frac12$$

And then we let $s = \cos(\varphi)$ which gives us:

$$\begin{array}{rcl} \cos(3\varphi) &=& -\dfrac12 \\ \cos(3\arccos(s)) &=& -\dfrac12 \\ \cos(3\arccos(2x)) &=& -\dfrac12 \\ \cos(3\arccos(2t-1)) &=& -\dfrac12 \\ \cos(3\arccos(2u^2-1)) &=& -\dfrac12 \\ \cos(3\arccos(2(\sin\theta)^2-1)) &=& -\dfrac12 \\ \cos(3\arccos(-\cos(2\theta))) &=& -\dfrac12 \\ \cos(3(\pi-2\theta)) &=& -\dfrac12 \\ \cos(6\theta) &=& \dfrac12 \\ \end{array}$$

Answer 5

Using http://mathworld.wolfram.com/WernerFormulas.html,

$$2\sin t(2\sin2t\sin4t)=2\sin t(\cos2t-\cos6t)$$

$$\sin3t-\sin t-2\sin t\cos6t$$

So, we have $$\sin t(1+2\cos6t)=0$$

Answer 6

$$\sin (3\theta) = 4\sin(\theta)\sin(2\theta) \sin(4\theta) \qquad \theta \in (0, \pi)$$

\begin{align} \sin (3\theta) &= 4\sin(\theta)\sin(2\theta) \sin(4\theta) \\ \cos(\theta)\sin(3\theta) &= 4\cos(\theta)\sin(\theta)\sin(2\theta)\sin(4\theta) \\ \frac 12 \sin(2\theta) + \frac 12 \sin(4 \theta) &= 2\sin(2\theta)\sin(2\theta)\sin(4\theta) \\ \frac 12 \sin(2\theta) + \sin(2\theta) \cos(2\theta) &= 2\sin(2\theta)\sin(2\theta)\sin(4\theta) \tag{A.}\\ \frac 12 + \cos(2\theta) &= 2\sin(2\theta)\sin(4\theta) \\ \frac 12 + \cos(2\theta) &= 4\sin^2(2\theta)\cos(2\theta) \\ 1 + 2\cos(2\theta) &= 8\cos(2\theta) - 8\cos^3(2\theta) \\ 8\cos^3(2\theta) - 6\cos(2\theta) &= -1 \tag{B.}\\ \cos(6\theta) &= -\frac 12 \\ \theta &\in \left\{\frac \pi9, \frac{2\pi}9, \frac{4\pi}9, \frac{5\pi}9, \frac{7\pi}9, \right\} \end{align}

Notes:

A. In the interval $(0,\pi)$, if $\sin(2\theta)=0$, then $\sin(3\theta) \ne 0$. So we can assume that $\sin(2\theta)\ne 0$.

B. $\cos(3x) = 4\cos^3(x) - 3\cos(x)$