Find the general solution of ' $\sin x+\sin 2 x$ $+\sin 3 x=\cos x+\cos 2 x+\cos 3 x^{\prime}$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \sin x+\sin 2 x+\sin 3 x=\cos x+\cos 2 x+\cos 3 x \ & (\sin x+\sin 3 x)+\sin 2 x=(\cos x+\cos 3 x)+\cos 2 x \ & 2 \sin \left(\frac{4 x}{2}\right) \cdot \cos \left(\frac{2 x}{2}\right)+\sin 2 x=2 \cos \left(\frac{4 x}{2}\right) \cos \left(\frac{2 x}{2}\right)+\cos 2 x \ & 2 \sin 2 x \cos x+\sin 2 x=2 \cos 2 x \cdot \cos x+\cos 2 x \ & \sin 2 x(2 \cos x+1)=\cos 2 x(2 \cos x+1) \ & (\sin 2 x-\cos 2 x)(2 \cos x+1)=0 \ & \sin 2 x-\cos 2 x=0 \text { (or) } 2 \cos x+1=0 \ & \tan 2 x=1=\tan \frac{\pi}{4} \ & \Rightarrow 2 x=n \pi+\frac{\pi}{4} \Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{8} \ & \text {or } \quad \cos x=\frac{-1}{2} \ & \cos x=\cos \frac{2 \pi}{3} \ & x=2 n \pi \pm \frac{2 \pi}{3} \ \end{aligned}$ Hence, option (d) is correct.

Find the general solution of ' \(\sin x+\sin 2 x\) \(+\sin 3 x=\cos x+\cos 2 x+\cos 3 x^{\prime}\)

Solution