sin 3theta = 4 sin, theta ,sin ,2theta ,sin , | vedclass

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Question

given equation can be written as

$3 \,sin\, 	heta - 4sin^3	heta = 4sin	heta \,sin\, 2	heta \,sin4	heta$

hence either $sin\, 	heta = 0 \Rig

Vedclass · Accepted Answer

$8$ real solutions.

Vedclass · Answer

given equation can be written as

$3 \,sin\, 	heta - 4sin^3	heta = 4sin	heta \,sin\, 2	heta \,sin4	heta$

hence either $sin\, 	heta = 0 \Rightarrow\, 	heta = n\pi$

or        $3 - 4sin^2	heta = 4\, sin\, 2	heta \,sin \,4	heta$

           $3 - 2 (1 - cos\, 2	heta ) = 2 (cos\, 2	heta - cos \,6	heta )$

or        $1 = - 2 cos\, 6	heta$

            $cos \,6	heta = - \frac{1}{2} = cos\, \frac{2 \pi}{3}$

            $6	heta = 2n \pi &plusmn; \frac{2 \pi}{3}$

if $0\, \le\, 	heta\, \le\, \pi$ then total solution are $0,\frac{\pi }{9},\frac{{2\pi }}{9},\frac{{4\pi }}{9},\frac{{5\pi }}{9},\frac{{7\pi }}{9},\frac{{8\pi }}{9},\pi $ is $8$ real solutions.

$sin 3\theta = 4 sin\, \theta \,sin \,2\theta \,sin \,4\theta$ in $0\, \le \,\theta\, \le \, \pi$ has :

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