Question

Solve the following equations:
(i) $\cos x+\cos 2x+\cos 3x=0$
(ii) $\cos x+\cos 3x-\cos 2x=0$
(iii) $\sin x+\sin 5x=\sin 3x$
(iv) $\cos x\cos 2x\cos 3x=\frac{1}{4}$
(v) $\cos x+\sin x=\cos 2x+\sin 2x$
(vi) $\sin x+\sin 2x+\sin 3=0$
(vii) $\sin x+\sin 2x+\sin 3x+\sin 4x=0$
(viii) $\sin 3x-\sin x=4{\cos }^{2}x-2$
(ix) $\sin 2x-\sin 4x+\sin 6x=0$

Answer 1

(i) $\cos x+\cos 2x+\cos 3x=0$
Now,
$$\begin{gathered} (\cos x+\cos 3x)+\cos 2x=0 \\ \Rightarrow 2\cos \left(\frac{4x}{2}\right)\cos \left(\frac{2x}{2}\right)+\cos 2x=0 \\ \Rightarrow 2\cos 2x\cos x+\cos 2x=0 \\ \Rightarrow \cos 2x(2\cos x+1)=0 \end{gathered}$$
$\Rightarrow \cos 2x=0$ or, $2\cos x+1=0$
$\Rightarrow \cos 2x=\cos \frac{\mathrm{\pi }}{2}$ or $\cos x=-\frac{1}{2}=\cos \frac{2\mathrm{\pi }}{3}$
$\Rightarrow 2x=(2n+1)\frac{\mathrm{\pi }}{2}$, $n\in Z$ or $x=2m\pi \pm \frac{2\mathrm{\pi }}{3},m\in Z$
$\Rightarrow x=(2n+1)\frac{\mathrm{\pi }}{4},n\in Z$ or $x=2m\pi \pm \frac{2\mathrm{\pi }}{3},m\in Z$

(ii) $(\cos x+\cos 3x)-\cos 2x=0$
$$\begin{gathered} \Rightarrow 2\cos \left(\frac{4x}{2}\right)\cos \left(\frac{2x}{2}\right)-\cos 2x=0 \\ \Rightarrow 2\cos 2x\cos x-\cos 2x=0 \\ \Rightarrow \cos 2x(2\cos x-1)=0 \end{gathered}$$

$\Rightarrow \cos 2x=0$ or $2\cos x-1=0$
$\Rightarrow \cos 2x=\cos \frac{\mathrm{\pi }}{2}$ or $\cos x=\frac{1}{2}\Rightarrow \cos x=\cos \frac{\mathrm{\pi }}{3}$
$\Rightarrow 2x=(2n+1)\frac{\mathrm{\pi }}{2},n\in Z$ or $x=2m\pi \pm \frac{\mathrm{\pi }}{3},m\in Z$
$\Rightarrow x=(2n+1)\frac{\mathrm{\pi }}{4},n\in Z$ or $x=2m\pi \pm \frac{\mathrm{\pi }}{3},m\in Z$

(iii) $\sin x+\sin 5x=\sin 3x$
$$\begin{gathered} \Rightarrow 2\sin \left(\frac{6x}{2}\right)\cos \left(\frac{4x}{2}\right)=\sin 3x \\ \Rightarrow 2\sin 3x\cos 2x=\sin 3x \\ \Rightarrow 2\sin 3x\cos 2x-\sin 3x=0 \\ \Rightarrow \sin 3x(2\cos 2x-1)=0 \end{gathered}$$

$\Rightarrow \sin 3x=0$ or $(2\cos 2x-1)=0$
$\Rightarrow \sin 3x=\sin 0$ or $\cos 2x=\frac{1}{2}=\cos \frac{\mathrm{\pi }}{3}$
$\Rightarrow 3x=n\pi$ or $2x=2m\pi \pm \frac{\mathrm{\pi }}{3}$
$\Rightarrow$ $x=\frac{n\mathrm{\pi }}{3},n\in Z$ or $x=m\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6},m\in \mathit{}Z$

$$\begin{gathered} \left(\mathrm{iv}\right)\cos x\cos 2x\cos 3x=\frac{1}{4} \\ \Rightarrow \left[\frac{\cos \left(x+2x\right)+\cos \left(2x-x\right)}{2}\right]\cos 3x=\frac{1}{4} \\ \Rightarrow 2\left[\cos 3x+\cos x\right]\cos 3x=1 \\ \Rightarrow 2{\cos }^{2}3x+2\cos x\cos 3x-1=0 \\ \Rightarrow 2{\cos }^{2}3x-1+2\cos x\cos 3x=0 \\ \Rightarrow \cos 6x+\cos 4x+\cos 2x=0 \\ \Rightarrow \cos 6x+\cos 2x+\cos 4x=0 \\ \Rightarrow 2\cos 4xc\mathrm{os}2x+\cos 4x=0 \\ \Rightarrow \cos 4x\left(2\cos 2x+1\right)=0 \\ \Rightarrow \cos 4x=0\mathrm{or}2\cos 2x+1=0 \\ \Rightarrow \cos 4x=0\mathrm{or}\cos 2x=\frac{-1}{2} \\ \Rightarrow \cos 4x=\cos \frac{\mathrm{\pi }}{2}\mathrm{or}\cos 2\mathrm{x}=\cos \frac{2\mathrm{\pi }}{3} \\ \Rightarrow 4x=\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in Z\mathrm{or}2x=2m\mathrm{\pi }\pm \frac{2\mathrm{\pi }}{3},m\in Z \\ \Rightarrow x=\left(2n+1\right)\frac{\mathrm{\pi }}{8},n\in Z\mathrm{or}x=m\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3},m\in Z \end{gathered}$$

(v) $\cos x+\sin x=\cos 2x+\sin 2x$

$$\begin{gathered} \Rightarrow \cos x-\cos 2x=\sin 2x-\sin x \\ \Rightarrow -2\sin \left(\frac{3x}{2}\right)\sin \left(\frac{-x}{2}\right)=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right) \\ \Rightarrow 2\sin \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{3x}{2}\right) \\ \Rightarrow 2\sin \left(\frac{x}{2}\right)\left[\sin \left(\frac{3x}{2}\right)-\cos \left(\frac{3x}{2}\right)\right]=0 \end{gathered}$$

$\Rightarrow \sin \frac{x}{2}=0$ or $\sin \frac{3x}{2}-\cos \frac{3x}{2}=0$
$\Rightarrow \sin \frac{x}{2}=\sin 0$ or $\sin \frac{3x}{2}=\cos \frac{3x}{2}$
$\Rightarrow$ $\frac{x}{2}=n\pi$, $n\in Z$ or $\cos \frac{3x}{2}=\cos \left(\frac{\mathrm{\pi }}{2}-\frac{3x}{2}\right)$
$\Rightarrow x=2n\pi ,n\in Z$ or $\frac{3x}{2}=2m\pi \pm \left(\frac{\mathrm{\pi }}{2}-\frac{3x}{2}\right),m\in Z$
$\Rightarrow$$x=2n\pi ,n\in Z$ or $\frac{3x}{2}=2m\pi +\frac{\mathrm{\pi }}{2}-\frac{3x}{2},m\in Z$ (Taking negative sign will give absurd result.)
$x=2n\pi ,n\in Z$ or $x=\frac{2m\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6},m\in Z$

(vi) $\sin x+\sin 2x+\sin 3x=0$

$$\begin{gathered} \Rightarrow \sin x+\sin 3x+\sin 2x=0 \\ \Rightarrow 2\sin \left(\frac{4x}{2}\right)\cos \left(\frac{2x}{2}\right)+\sin 2x=0 \\ \Rightarrow 2\sin 2x\cos x+\sin 2x=0 \\ \Rightarrow \sin 2x(2\cos x+1)=0 \end{gathered}$$

$\Rightarrow \sin 2x=0$ or $2\cos x+1=0$
$\Rightarrow \sin 2x=\sin 0$ or $\cos x=-\frac{1}{2}\Rightarrow \cos x=\cos \frac{2\mathrm{\pi }}{3}$
$\Rightarrow$$x=\frac{n\mathrm{\pi }}{2},n\in Z$ or $x=2m\pi \pm \frac{2\mathrm{\pi }}{3}$, $m\in Z$

(vii) $\sin x+\sin 2x+\sin 3x+\sin 4x=0$

$$\begin{gathered} \Rightarrow \sin 3x+\sin x+\sin 4x+\sin 2x=0 \\ \Rightarrow 2\sin \left(\frac{4x}{2}\right)\cos \left(\frac{2x}{2}\right)+2\sin \left(\frac{6x}{2}\right)\cos \left(\frac{2x}{2}\right)=0 \\ \Rightarrow 2\sin 2x\cos x+2\sin 3x\cos x=0 \\ \Rightarrow 2\cos x(\sin 2x+\sin 3x)=0 \\ \Rightarrow 2\cos x\left(2\sin \left(\frac{5x}{2}\right)\cos \left(\frac{x}{2}\right)\right)=0 \\ \Rightarrow 4\cos x\sin \left(\frac{5x}{2}\right)\cos \left(\frac{x}{2}\right)=0 \end{gathered}$$
$\Rightarrow \cos x=0,\sin \left(\frac{5x}{2}\right)=0$ or $\cos \left(\frac{x}{2}\right)=0$
$\Rightarrow \cos x=\cos \frac{\mathrm{\pi }}{2},\sin \left(\frac{5x}{2}\right)=\sin 0$ or $\cos \left(\frac{x}{2}\right)=\cos \frac{\mathrm{\pi }}{2}$
$\Rightarrow x=(2n+1)\frac{\mathrm{\pi }}{2},n\in Zor\frac{5x}{2}=n\pi ,n\in Z$ or, $\frac{x}{2}=(2n+1)\frac{\mathrm{\pi }}{2},n\in Z$
$\Rightarrow x=(2n+1)\frac{\mathrm{\pi }}{2},n\in Z$ or $x=\frac{2n\mathrm{\pi }}{5},n\in Z$ or $x=(2n+1)\pi ,n\in Z$

(viii) $\sin 3x-\sin x=4{\cos }^{2}x-2$
$$\begin{gathered} \Rightarrow \sin 3x-\sin x=2(2{\cos }^{2}x-1) \\ \Rightarrow 2\sin \left(\frac{2x}{2}\right)\cos \left(\frac{4x}{2}\right)=2\cos 2x \\ \Rightarrow 2\sin x\cos 2x=2\cos 2x \\ \Rightarrow \sin x\cos 2x=\cos 2x \\ \Rightarrow \cos 2x(\sin x-1)=0 \end{gathered}$$
$\Rightarrow \cos 2x=0$ or $\sin x-1=0$
$\Rightarrow$$\cos 2x=\cos \frac{\mathrm{\pi }}{2}$ or $\sin x=1\Rightarrow \sin x=\sin \frac{\mathrm{\pi }}{2}$
$\Rightarrow 2x=(2n+1)\frac{\mathrm{\pi }}{2}$, $n\in Z$ or $x=n\pi +(-1{)}^n\frac{\mathrm{\pi }}{2},n\in Z$
$\Rightarrow x=(2n\hspace{0.17em}+1)\frac{\mathrm{\pi }}{4},n\in Z$ or $x=n\pi +(-1{)}^n\frac{\mathrm{\pi }}{2},n\in Z$

(ix) $\sin 2x-\sin 4x+\sin 6x=0.$
$$\begin{gathered} \Rightarrow 2\sin \left(\frac{8x}{2}\right)\cos \left(\frac{4x}{2}\right)-\sin 4x=0 \\ \Rightarrow 2\sin 4x\cos 2x-\sin 4x=0 \\ \Rightarrow \sin 4x(2\cos 2x-1)=0 \end{gathered}$$
$\Rightarrow \sin 4x=0$ or $2\cos 2x-1=0$
$\Rightarrow 4x=n\pi$, $n\in Z$ or $\cos 2x=\frac{1}{2}\Rightarrow \cos 2x=\cos \frac{\mathrm{\pi }}{3}$
$\Rightarrow x=\frac{n\mathrm{\pi }}{4},n\in Z$ or $x=n\pi \pm \frac{\mathrm{\pi }}{6},n\in Z$