How do you solve $2 sin x + 1 = 0$ $2 sin x + 1 = 0$ ?

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How do you solve $2 sin x + 1 = 0$ $2 sin x + 1 = 0$ ?

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Answer 1 · 2015-05-02T22:28:41

Subtract 1, divide by 2.

$sin x = - \frac{1}{2}$ $sinx = -1/2$

$sin x = \pm \frac{1}{2}$ $sinx = pm1/2$ at $30^{o}$ $30^o$ , $150^{o}$ $150^o$ , $210^{o}$ $210^o$ , and $330^{o}$ $330^o$ .
$sin x = - \frac{1}{2}$ $sinx = -1/2$ at $210^{o}$ $210^o$ and $330^{o}$ $330^o$ .

So, you have two answers. Really though, you have infinite answers if you consider that there are equivalent angles at every $\pm 360^{o}$ $pm360^o$ . What you get, then, is:

$\frac{210^{o}}{180^{o}} \cdot π = \frac{7 π}{6}$ $210^o/180^o*pi = (7pi)/6$

$\frac{330^{o}}{180^{o}} \cdot π = \frac{11 π}{6}$ $330^o/180^o*pi = (11pi)/6$

Therefore, $sin x = - \frac{1}{2}$ $sinx = -1/2$ at $\frac{7 π}{6}$ $(7pi)/6$ , $\frac{11 π}{6}$ $(11pi)/6$ , $\frac{19 π}{6}$ $(19pi)/6$ , $\frac{23 π}{6}$ $(23pi)/6$ , etc.

Notice how $\frac{11 π}{6}$ $(11pi)/6$ = $2 π - \frac{π}{6}$ $2pi - pi/6$ , and $\frac{7 π}{6} = 0 + \frac{7 n π}{6}$ $(7pi)/6 = 0 + (7npi)/6$ . Recall how $sin 0 = sin 2 π = sin 2 n π$ $sin 0 = sin 2pi = sin 2npi$ , where n $in ZZ$ (n is in the set of integers). Thus, you can write:

$x = 2npi-pi/6$

$x = 2npi+(7pi)/6$

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