Let theta be an acute angle such that the equ | vedclass

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(C) The given equation is $x^3+4 x^2 \cos 	heta+x \cot 	heta=0$. Factoring out $x$,we get $x(x^2+4 x \cos 	heta+\cot 	heta)=0$. One root is $x=0$.

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

Vedclass · Answer

(C) The given equation is $x^3+4 x^2 \cos 	heta+x \cot 	heta=0$. Factoring out $x$,we get $x(x^2+4 x \cos 	heta+\cot 	heta)=0$. One root is $x=0$. For the equation to have multiple roots,either the quadratic part $x^2+4 x \cos 	heta+\cot 	heta=0$ has equal roots (discriminant $D=0$) or $x=0$ is a root of the quadratic part. Case $1$: $D = (4 \cos 	heta)^2 - 4(1)(\cot 	heta) = 0$. $16 \cos^2 	heta - 4 \cot 	heta = 0 \Rightarrow 4 \cos^2 	heta = \frac{\cos 	heta}{\sin 	heta}$. This implies $\cos 	heta = 0$ (not possible as $	heta$ is acute) or $4 \cos 	heta \sin 	heta = 1$. $2 \sin 2 	heta = 1 \Rightarrow \sin 2 	heta = \frac{1}{2}$. $2 	heta = \frac{\pi}{6}, \frac{5 \pi}{6} \Rightarrow 	heta = \frac{\pi}{12}, \frac{5 \pi}{12}$. Case $2$: $x=0$ is a root of $x^2+4 x \cos 	heta+\cot 	heta=0$,which implies $\cot 	heta = 0$,so $	heta = \frac{\pi}{2}$ (not acute). Thus,the values are $	heta = \frac{\pi}{12} 	ext{ or } \frac{5 \pi}{12}$.

Let $\theta$ be an acute angle such that the equation $x^3+4 x^2 \cos \theta+x \cot \theta=0$ has multiple roots. Then the value of $\theta$ (in radians) is

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