How do you solve the following equation $cos x (2 sin x + 1) = 0$ $cosx(2sinx+1)=0$ in the interval [0, 2pi]?

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How do you solve the following equation $cos x (2 sin x + 1) = 0$ $cosx(2sinx+1)=0$ in the interval [0, 2pi]?

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Answer 1 · 2016-10-02T15:23:04

$The Soln. Set = {\frac{π}{2}, 7 \frac{π}{6}, 3 \frac{π}{2}, 11 \frac{π}{6}} \subset [0, 2 π]$ $"The Soln. Set "={pi/2, 7pi/6, 3pi/2, 11pi/6} sub [0,2pi]$ .

Explanation:

$cos x (2 sin x + 1) = 0$ $cosx(2sinx+1)=0$

$\Rightarrow cos x = 0, sin x = - \frac{1}{2}$ $rArr cosx=0, sinx=-1/2$

$cos x = 0, and, 0 \leq x \leq 2 π \Rightarrow x = \frac{π}{2}, 3 \frac{π}{2} .$ $cosx=0, and, 0 le x le 2pi rArr x=pi/2, 3pi/2.$

$sin x = - \frac{1}{2}, &, x \in [0, 2 π] \Rightarrow x = π + \frac{π}{6}, or, x = 2 π - \frac{π}{6},$ $sinx=-1/2, &, x in [0,2pi] rArr x=pi+pi/6, or, x=2pi-pi/6,$

$i.e., x = 7 \frac{π}{6}, 11 \frac{π}{6} .$ $"i.e., "x=7pi/6, 11pi/6.$

Thus, the Soln. Set $= {\frac{π}{2}, 7 \frac{π}{6}, 3 \frac{π}{2}, 11 \frac{π}{6}} \subset [0, 2 π]$ $={pi/2, 7pi/6, 3pi/2, 11pi/6} sub [0,2pi]$ .

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How do you solve the following equation $cos x (2 sin x + 1) = 0$ $cosx(2sinx+1)=0$ in the interval [0, 2pi]?

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Explanation:

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