If cos(ptheta) + cos(qtheta) = 0 and p neq q, | vedclass

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(B) Given $\cos(p	heta) + \cos(q	heta) = 0$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$2 \cos(\

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$\frac{(2n + 1)\pi}{p + q}$ or $\frac{(2n + 1)\pi}{p - q}$

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(B) Given $\cos(p	heta) + \cos(q	heta) = 0$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$2 \cos(\frac{(p+q)	heta}{2}) \cos(\frac{(p-q)	heta}{2}) = 0$.This implies either $\cos(\frac{(p+q)	heta}{2}) = 0$ or $\cos(\frac{(p-q)	heta}{2}) = 0$.For $\cos x = 0$, the general solution is $x = (2n+1)\frac{\pi}{2}$.Case $1$: $\frac{(p+q)	heta}{2} = (2n+1)\frac{\pi}{2} \implies 	heta = \frac{(2n+1)\pi}{p+q}$.Case $2$: $\frac{(p-q)	heta}{2} = (2n+1)\frac{\pi}{2} \implies 	heta = \frac{(2n+1)\pi}{p-q}$.Thus, $	heta = \frac{(2n+1)\pi}{p \pm q}$.

If $\cos(p\theta) + \cos(q\theta) = 0$ and $p \neq q$, then the general value of $\theta$ (where $n$ is any integer) is...

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