Let theta, 0

Question

(b)

$x^2+4 x \cos 	heta+\cot 	heta=0$

$D =0 \Rightarrow 16 \cos ^2 	heta=4 \cot 	heta$

$\Rightarrow 4 \cos ^2 	heta=\frac{\cos 	heta}{\

Vedclass · Accepted Answer

$\frac{\pi}{12}$ or $\frac{5 \pi}{12}$

Vedclass · Answer

(b)

$x^2+4 x \cos 	heta+\cot 	heta=0$

$D =0 \Rightarrow 16 \cos ^2 	heta=4 \cot 	heta$

$\Rightarrow 4 \cos ^2 	heta=\frac{\cos 	heta}{\sin 	heta}$

$\Rightarrow \sin 2 	heta=\frac{1}{2}$

$\Rightarrow 2 	heta=\frac{\pi}{6}, \frac{5 \pi}{6}$

$\Rightarrow 	heta=\frac{\pi}{12}, \frac{5 \pi}{12}$

Let $\theta, 0 < \theta < \pi / 2$, be an angle such that the equation $x ^2+4 x \cos \theta+\cot \theta=0$ has equal roots for $x$. Then $\theta$ in radians is

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