# Solve the following equations:

Question:

Solve the following equations:

(i) $\cos x+\cos 2 x+\cos 3 x=0$

(ii) $\cos x+\cos 3 x-\cos 2 x=0$

(iii) $\sin x+\sin 5 x=\sin 3 x$

(iv) $\cos x \cos 2 x \cos 3 x=\frac{1}{4}$

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

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

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

(viii) $\sin 3 x-\sin x=4 \cos ^{2} x-2$

(ix) $\sin 2 x-\sin 4 x+\sin 6 x=0$

Solution:

(i) $\cos x+\cos 2 x+\cos 3 x=0$

Now,

$(\cos x+\cos 3 x)+\cos 2 x=0$

$\Rightarrow 2 \cos \left(\frac{4 x}{2}\right) \cos \left(\frac{2 x}{2}\right)+\cos 2 x=0$

$\Rightarrow 2 \cos 2 x \cos x+\cos 2 x=0$

$\Rightarrow \cos 2 x(2 \cos x+1)=0$

$\Rightarrow \cos 2 x=0$ or, $2 \cos x+1=0$

$\Rightarrow \cos 2 x=\cos \frac{\pi}{2}$ or $\cos x=-\frac{1}{2}=\cos \frac{2 \pi}{3}$

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

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

(ii) $(\cos x+\cos 3 x)-\cos 2 x=0$

$\Rightarrow 2 \cos \left(\frac{4 x}{2}\right) \cos \left(\frac{2 x}{2}\right)-\cos 2 x=0$

$\Rightarrow 2 \cos 2 x \cos x-\cos 2 x=0$

$\Rightarrow \cos 2 x(2 \cos x-1)=0$

$\Rightarrow \cos 2 x=0$ or $2 \cos x-1=0$

$\Rightarrow \cos 2 x=\cos \frac{\pi}{2}$ or $\cos x=\frac{1}{2} \Rightarrow \cos x=\cos \frac{\pi}{3}$

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

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

(iii) $\sin x+\sin 5 x=\sin 3 x$

$\Rightarrow 2 \sin \left(\frac{6 x}{2}\right) \cos \left(\frac{4 x}{2}\right)=\sin 3 x$

$\Rightarrow 2 \sin 3 x \cos 2 x=\sin 3 x$

$\Rightarrow 2 \sin 3 x \cos 2 x-\sin 3 x=0$

$\Rightarrow \sin 3 x(2 \cos 2 x-1)=0$

$\Rightarrow \sin 3 x=0$ or $(2 \cos 2 x-1)=0$

$\Rightarrow \sin 3 x=\sin 0$ or $\cos 2 x=\frac{1}{2}=\cos \frac{\pi}{3}$

$\Rightarrow 3 x=n \pi$ or $2 x=2 m \pi \pm \frac{\pi}{3}$

$\Rightarrow x=\frac{n \pi}{3}, n \in Z$ or $x=m \pi \pm \frac{\pi}{6}, m \in Z$

(iv) $\cos x \cos 2 x \cos 3 x=\frac{1}{4}$

$\Rightarrow\left[\frac{\cos (x+2 x)+\cos (2 x-x)}{2}\right] \cos 3 x=\frac{1}{4}$

$\Rightarrow 2[\cos 3 x+\cos x] \cos 3 x=1$

$\Rightarrow 2 \cos ^{2} 3 x+2 \cos x \cos 3 x-1=0$

$\Rightarrow 2 \cos ^{2} 3 x-1+2 \cos x \cos 3 x=0$

$\Rightarrow \cos 6 x+\cos 4 x+\cos 2 x=0$

$\Rightarrow \cos 6 x+\cos 2 x+\cos 4 x=0$

$\Rightarrow 2 \cos 4 x c \cos 2 x+\cos 4 x=0$

$\Rightarrow \cos 4 x(2 \cos 2 x+1)=0$

$\Rightarrow \cos 4 x=0$ or $2 \cos 2 x+1=0$

$\Rightarrow \cos 4 x=0$ or $\cos 2 x=\frac{-1}{2}$

$\Rightarrow \cos 4 x=\cos \frac{\pi}{2}$ or $\cos 2 \mathrm{x}=\cos \frac{2 \pi}{3}$

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

$\Rightarrow x=(2 n+1) \frac{\pi}{8}, n \in Z$ or $x=m \pi \pm \frac{\pi}{3}, m \in Z$

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

$\Rightarrow \cos x-\cos 2 x=\sin 2 x-\sin x$

$\Rightarrow-2 \sin \left(\frac{3 x}{2}\right) \sin \left(\frac{-x}{2}\right)=2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{3 x}{2}\right)$

$\Rightarrow 2 \sin \left(\frac{3 x}{2}\right) \sin \left(\frac{x}{2}\right)=2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{3 x}{2}\right)$

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

$\Rightarrow \sin \frac{x}{2}=0$ or $\sin \frac{3 x}{2}-\cos \frac{3 x}{2}=0$

$\Rightarrow \sin \frac{x}{2}=\sin 0$ or $\sin \frac{3 x}{2}=\cos \frac{3 x}{2}$

$\Rightarrow \frac{x}{2}=n \pi, n \in Z$ or $\cos \frac{3 x}{2}=\cos \left(\frac{\pi}{2}-\frac{3 x}{2}\right)$

$\Rightarrow x=2 n \pi, n \in Z$ or $\frac{3 x}{2}=2 m \pi \pm\left(\frac{\pi}{2}-\frac{3 x}{2}\right), m \in Z$

$\Rightarrow x=2 n \pi, n \in Z$ or $\frac{3 x}{2}=2 m \pi+\frac{\pi}{2}-\frac{3 x}{2}, m \in Z \quad$ (Taking negative sign will give absurd result.)

$x=2 n \pi, n \in Z$ or $x=\frac{2 m \pi}{3}+\frac{\pi}{6}, m \in Z$

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

$\Rightarrow \sin x+\sin 3 x+\sin 2 x=0$

$\Rightarrow 2 \sin \left(\frac{4 x}{2}\right) \cos \left(\frac{2 x}{2}\right)+\sin 2 x=0$

$\Rightarrow 2 \sin 2 x \cos x+\sin 2 x=0$

$\Rightarrow \sin 2 x(2 \cos x+1)=0$

$\Rightarrow \sin 2 x=0$ or $2 \cos x+1=0$

$\Rightarrow \sin 2 x=\sin 0$ or $\cos x=-\frac{1}{2} \Rightarrow \cos x=\cos \frac{2 \pi}{3}$

$\Rightarrow x=\frac{n \pi}{2}, n \in Z$ or $x=2 m \pi \pm \frac{2 \pi}{3}, m \in Z$

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

$\Rightarrow \sin 3 x+\sin x+\sin 4 x+\sin 2 x=0$

$\Rightarrow 2 \sin \left(\frac{4 x}{2}\right) \cos \left(\frac{2 x}{2}\right)+2 \sin \left(\frac{6 x}{2}\right) \cos \left(\frac{2 x}{2}\right)=0$

$\Rightarrow 2 \sin 2 x \cos x+2 \sin 3 x \cos x=0$

$\Rightarrow 2 \cos x(\sin 2 x+\sin 3 x)=0$

$\Rightarrow 2 \cos x\left(2 \sin \left(\frac{5 x}{2}\right) \cos \left(\frac{x}{2}\right)\right)=0$

$\Rightarrow 4 \cos x \sin \left(\frac{5 x}{2}\right) \cos \left(\frac{x}{2}\right)=0$

$\Rightarrow \cos x=0, \sin \left(\frac{5 x}{2}\right)=0$ or $\cos \left(\frac{x}{2}\right)=0$

$\Rightarrow \cos x=\cos \frac{\pi}{2}, \sin \left(\frac{5 x}{2}\right)=\sin 0$ or $\cos \left(\frac{x}{2}\right)=\cos \frac{\pi}{2}$

$\Rightarrow x=(2 n+1) \frac{\pi}{2}, n \in Z$ or $\frac{5 x}{2}=n \pi, n \in Z$ or, $\frac{x}{2}=(2 n+1) \frac{\pi}{2}, n \in Z$

$\Rightarrow x=(2 n+1) \frac{\pi}{2}, n \in Z$ or $x=\frac{2 n \pi}{5}, n \in Z$ or $x=(2 n+1) \pi, n \in Z$

(viii) $\sin 3 x-\sin x=4 \cos ^{2} x-2$

$\Rightarrow \sin 3 x-\sin x=2\left(2 \cos ^{2} x-1\right)$

$\Rightarrow 2 \sin \left(\frac{2 x}{2}\right) \cos \left(\frac{4 x}{2}\right)=2 \cos 2 x$

$\Rightarrow 2 \sin x \cos 2 x=2 \cos 2 x$

$\Rightarrow \sin x \cos 2 x=\cos 2 x$

$\Rightarrow \cos 2 x(\sin x-1)=0$

$\Rightarrow \cos 2 x=0$ or $\sin x-1=0$

$\Rightarrow \cos 2 x=\cos \frac{\pi}{2}$ or $\sin x=1 \Rightarrow \sin x=\sin \frac{\pi}{2}$

$\Rightarrow 2 x=(2 n+1) \frac{\pi}{2}, n \in Z$ or $x=n \pi+(-1)^{n} \frac{\pi}{2}, n \in Z$

$\Rightarrow x=(2 n+1) \frac{\pi}{4}, n \in Z$ or $x=n \pi+(-1)^{n} \frac{\pi}{2}, n \in Z$

(ix) $\sin 2 x-\sin 4 x+\sin 6 x=0$.

$\Rightarrow 2 \sin \left(\frac{8 x}{2}\right) \cos \left(\frac{4 x}{2}\right)-\sin 4 x=0$

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

$\Rightarrow \sin 4 x(2 \cos 2 x-1)=0$

$\Rightarrow \sin 4 x=0$ or $2 \cos 2 x-1=0$

$\Rightarrow 4 x=n \pi, n \in Z$ or $\cos 2 x=\frac{1}{2} \Rightarrow \cos 2 x=\cos \frac{\pi}{3}$

$\Rightarrow x=\frac{n \pi}{4}, n \in Z$ or $x=n \pi \pm \frac{\pi}{6}, n \in Z$