Question

If $\cos \left(\theta \right)+\cos \left(2\theta \right)+\cos \left(3\theta \right)=0,$ then the general value of $\theta$ is:

Answer 1

Find the general value of $\theta$:

$$\begin{gathered} \cos \left(\theta \right)+\cos \left(2\theta \right)+\cos \left(3\theta \right)=0 \\ \Rightarrow \left(\cos \left(\theta \right)+\cos \left(3\theta \right)\right)+\cos \left(2\theta \right)=0 \\ \Rightarrow 2\cos \left(2\theta \right)\cos \left(\theta \right)+\cos \left(2\theta \right)=0;\left[\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\right] \\ \Rightarrow \cos \left(2\theta \right)(2\cos \left(\theta \right)+1)=0 \end{gathered}$$

Either $\cos \left(2\theta \right)=0$ or $2\cos \left(\theta \right)+1=0$

If,

$$\begin{gathered} \cos \left(2\theta \right)=0 \\ \Rightarrow 2\theta =2n\pi \pm \frac{\pi }{2} \\ \Rightarrow \theta =n\pi \pm \frac{\pi }{4} \end{gathered}$$ for $n=0,1,2,...$

If,

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

$\Rightarrow \theta =(2n+1)\pi \pm \frac{2\pi }{3}$ for $n=0,1,2,...$

Hence , value of $\theta$ is $n\pi \pm \frac{\pi }{4}$ or $(2n+1)\pi \pm \frac{2\pi }{3}$for $n=0,1,2,...$