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If $3\sin 2\theta=2\sin 3\theta$ and $0\;<\;\theta\;<\pi$ then value of $\sin\theta$ is

$(a)\;\large\frac{\sqrt 2}{3}$$\qquad(b)\;\large\frac{\sqrt 3}{\sqrt 5}$$\qquad(c)\;\large\frac{\sqrt {15}}{4}$$\qquad(d)\;\large\frac{\sqrt 2}{\sqrt 5}$

1 Answer

$3\sin 2\theta=2\sin 3\theta$
$\Rightarrow 3\sin\theta\cos\theta-3\sin\theta+4\sin^3\theta=0$
$\Rightarrow \sin\theta[3\cos\theta-3+4(1-\cos^2\theta)]=0$
$\Rightarrow 4\cos^2\theta-3\cos\theta-1=0$
$(\cos\theta-1)(4\cos\theta+1)=0$
$4\cos\theta+1=0$
$4\cos\theta=-1$
$\cos\theta=\large\frac{-1}{4}$
$\sin\theta=\large\frac{\sqrt{15}}{4}$
Hence (c) is the correct answer.
answered Oct 8, 2013 by sreemathi.v
 

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