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Exercise 3.12 · Q12

Q.If cos⁡pθ+cos⁡qθ=0\cos p\theta+\cos q\theta=0 and if p≠qp\ne q, then θ\theta is equal to (nn is any integer)

(1) π(3n+1)p−q\dfrac{\pi(3n+1)}{p-q}
(2) π(2n+1)p±q\dfrac{\pi(2n+1)}{p\pm q}
(3) π(n±1)p±q\dfrac{\pi(n\pm1)}{p\pm q}
(4) π(n+2)p+q\dfrac{\pi(n+2)}{p+q}
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Write cos⁡pθ=−cos⁡qθ=cos⁡(π−qθ)\cos p\theta=-\cos q\theta=\cos(\pi-q\theta) and apply the general solution of cos⁡A=cos⁡B\cos A=\cos B; combine the ++ and −- branches.

Step 1. cos⁡pθ+cos⁡qθ=0⇒cos⁡pθ=−cos⁡qθ=cos⁡(π−qθ)\cos p\theta+\cos q\theta=0\Rightarrow\cos p\theta=-\cos q\theta=\cos(\pi-q\theta).

Step 2. General solution of cos⁡A=cos⁡B\cos A=\cos B is A=2mπ±BA=2m\pi\pm B (using mm here to avoid clashing with nn): pθ=2mπ±(π−qθ)p\theta=2m\pi\pm(\pi-q\theta).

Step 3. (++ branch). pθ=2mπ+π−qθ⇒(p+q)θ=(2m+1)π⇒θ=(2m+1)πp+qp\theta=2m\pi+\pi-q\theta\Rightarrow(p+q)\theta=(2m+1)\pi\Rightarrow\theta=\dfrac{(2m+1)\pi}{p+q}. …

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