One of solutions of equation 8 sin ^3 θ-7 sin θ+√{3} cos θ=0 lies in interval

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Trigonometrical Equations

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One of the solutions of the equation $8 \sin ^3 \theta-7 \sin \theta+\sqrt{3} \cos \theta=0$ lies in the interval

$\left(0^{\circ}, 10^{\circ}\right]$

$\left(10^{\circ}, 20^{\circ}\right)$

$\left(20^{\circ}, 30^{\circ}\right)$

$\left(30^{\circ}, 40^{\circ}\right]$

(KVPY-2017)

Solution

(b)

Given,

$8 \sin ^3 \theta-7 \sin \theta+\sqrt{3} \cos \theta =0$

$2\left(4 \sin ^3 \theta\right)-7 \sin \theta+\sqrt{3} \cos \theta =0$

$2(3 \sin \theta-\sin 3 \theta)-7 \sin \theta$

$+\sqrt{3} \cos \theta =0$

$6 \sin \theta-2 \sin 3 \theta-7 \sin \theta$

$+\sqrt{3} \cos \theta =0$

$\sqrt{3} \cos \theta-\sin \theta-2 \sin 3 \theta =0$

$\frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta =\sin 3 \theta$

$\sin \left(\frac{\pi}{3}-\theta\right) =\sin 3 \theta$

$\frac{\pi}{3}-\theta=3 \theta \Rightarrow 4 \theta=60^{\circ} \Rightarrow \theta =15^{\circ}$

$\theta \in\left(10^{\circ}, 20^{\circ}\right]$

Standard 11

Mathematics

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(KVPY-2011)

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One of the solutions of the equation $8 \sin ^3 \theta-7 \sin \theta+\sqrt{3} \cos \theta=0$ lies in the interval

Solution

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Let $S\, = \,\left\{ {\theta \, \in \,[ – \,2\,\pi ,\,\,2\,\pi ]\, :\,2\,{{\cos }^2}\,\theta \, + \,3\,\sin \,\theta \, = \,0} \right\}$. Then the sum of the elements of $S$ is

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Let $f(x)=\cos 5 x+A \cos 4 x+B \cos 3 x$ $+C \cos 2 x+D \cos x+E$, and

$T=f(0)-f\left(\frac{\pi}{5}\right)+f\left(\frac{2 \pi}{5}\right)-f\left(\frac{3 \pi}{5}\right)+\ldots+f\left(\frac{8 \pi}{5}\right)-f\left(\frac{9 \pi}{5}\right) \text {. }$Then, $T$