The general solution of the equation cot thet | vedclass

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Question

(D) Given equation: $\cot 	heta \cdot \cot 2	heta = 1$$\implies \frac{\cos 	heta}{\sin 	heta} \cdot \frac{\cos 2	heta}{\sin 2	heta} = 1$$\implie

Vedclass · Accepted Answer

$	heta = (2n+1)\frac{\pi}{6}, n \in Z$

Vedclass · Answer

(D) Given equation: $\cot 	heta \cdot \cot 2	heta = 1$$\implies \frac{\cos 	heta}{\sin 	heta} \cdot \frac{\cos 2	heta}{\sin 2	heta} = 1$$\implies \cos 	heta \cdot \cos 2	heta = \sin 	heta \cdot \sin 2	heta$$\implies \cos 	heta \cdot \cos 2	heta - \sin 	heta \cdot \sin 2	heta = 0$Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:$\implies \cos(	heta + 2	heta) = 0$$\implies \cos 3	heta = 0$The general solution for $\cos x = 0$ is $x = (2n+1)\frac{\pi}{2}, n \in Z$.Therefore, $3	heta = (2n+1)\frac{\pi}{2}$$\implies 	heta = (2n+1)\frac{\pi}{6}, n \in Z$.

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ has two elements
$(II)$ $\{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$	$(Q)$ has three elements
$(III)$ $\{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos (2 x)=\sqrt{3}\}$	$(R)$ has four elements
$(IV)$ $\{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$	$(S)$ has five elements
	$(T)$ has six elements

The general solution of the equation $\cot \theta \cdot \cot 2\theta = 1$ is...

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