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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The number of all possible values of $\theta$ where $0 < \theta < \pi$ for which the system of equations $(y+z)\cos 3\theta=(xyz)\sin 3\theta$,$x\sin 3\theta=\large\frac{2\cos 3\theta}{y}+\frac{2\sin 3\theta}{z}$,$(xyz)\sin 3\theta=(y+2z)\cos 3\theta+y\sin 3\theta$ have a solution $(x_0,y_0,z_0)$ with $y_0,z_0\neq 0$ is

$\begin{array}{1 1}(a)\;4&(b)\;5\\(c)\;2&(d)\;3\end{array}$

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1 Answer

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Step 1:
Given :
$(y+z)\cos 3\theta=(xyz)\sin 3\theta$
$\Rightarrow\:(xyz)\sin 3\theta=y\cos 3\theta+z\cos 3\theta$-----(1)
$\Rightarrow\:x\sin 3\theta=\large\frac{2\cos 3\theta}{y}+\frac{2\sin 3\theta}{z}$
$(xyz)\sin 3\theta=2y\sin 3\theta+2z\cos 3\theta$---------(2)
$(xyz)\sin 3\theta=y(\sin 3\theta+\cos 3\theta)+2z\cos 3\theta$-----(3)
From (1) & (2) we get,
$y\cos 3\theta+z\cos 3\theta=2y\sin 3\theta+2z\cos 3\theta$
$\Rightarrow\:(\cos 3\theta-2\sin 3\theta)y=z\cos 3\theta$-----(4)
From (1) & (3) we get
$y(\sin 3\theta+\cos 3\theta)+2z\cos 3\theta=y\cos 3\theta+z\cos 3\theta$
$\Rightarrow\:y\sin 3\theta+z\cos 3\theta=0$
$\Rightarrow\:z\cos 3\theta=-y\sin 3\theta$------(5)
Step 2:
From (4) and (5) we get,
$(\cos 3\theta-2\sin 3\theta)y=-y\sin 3\theta$
$\Rightarrow\:\cos 3\theta-2\sin 3\theta=-\sin 3\theta$
$\Rightarrow\:\cos 3\theta=\sin 3\theta$
$\Rightarrow\:\tan 3\theta=1=\tan\large\frac{\pi}{4}$
When $n=0,\pm 1,\pm 2$......
Values of $\theta$ for $0 <\theta <\pi$ are $\large\frac{\pi}{12},\frac{5\pi}{12},\frac{9\pi}{12}$
$\Rightarrow$ No. of values $\theta$ is 3
Hence (d) is the correct answer.
answered Nov 18, 2013 by sreemathi.v
edited Mar 25, 2014 by rvidyagovindarajan_1

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