Common roots of the equations 2{sin ^2}x + {s | vedclass

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Question

(b) $2{\sin ^2}x + {\sin ^2}2x = 2$ ......$(i)$

and $\sin 2x + \cos 2x = 	an x$.....$(ii)$

Solving $(i)$, ${\sin ^2}2x = 2{\cos ^2}x$

==>

Vedclass · Accepted Answer

$x = (2n + 1)\frac{\pi }{4}$

Vedclass · Answer

(b) $2{\sin ^2}x + {\sin ^2}2x = 2$ ......$(i)$

and $\sin 2x + \cos 2x = 	an x$.....$(ii)$

Solving $(i)$, ${\sin ^2}2x = 2{\cos ^2}x$

==>$2{\cos ^2}x\cos 2x = 0$

==>$x = (2n + 1)\frac{\pi }{2}{m{ or }}x = (2n + 1)\frac{\pi }{4}$

$	herefore $ Common roots are $(2n \pm 1)\frac{\pi }{4}$

Solving $(ii)$, $\frac{{2	an x + 1 - {{	an }^2}x}}{{1 + {{	an }^2}x}} = 	an x$

$ \Rightarrow $ ${	an ^3}x + {	an ^2}x - 	an x - 1 = 0$

$ \Rightarrow $ $({	an ^2}x - 1)\,(	an x + 1) = 0$

$ \Rightarrow $ $x = m\pi \pm \frac{\pi }{4}$

Trick : For $n = 0$, option $(a)$ gives $	heta = - \frac{\pi }{2}$ which satisfies the equation $(i)$ but does not satisfy the $(ii)$.

Now option $(b) $ gives $	heta = \frac{\pi }{4}$ which satisfies both the equations.

Common roots of the equations $2{\sin ^2}x + {\sin ^2}2x = 2$ and $\sin 2x + \cos 2x = \tan x,$ are

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