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

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

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Answer 1 · 2016-05-27T19:17:31

$x = \frac{11 π}{6}, \frac{7 π}{6}$ $x = (11pi)/6, (7pi)/6$

Explanation:

To solve this equation, go about it as you would any other equation. Get the sin x all by itself.

$2 sin x + 1 = 0$ $2 sin x +1 = 0$

$2 sin x = - 1$ $2 sin x = -1$

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

Then, use the unit circle to find all radian values which have a y-coordinate of $- \frac{1}{2}$ $-1/2$ , since sin is the $y$ $y$ value (as opposed to cos, which is the x value).

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As you can see, the coordinates $(- \frac{\sqrt{3}}{2}$ $(-sqrt(3)/2$ , $- \frac{1}{2})$ $-1/2)$ and $(\frac{\sqrt{3}}{2}, - \frac{1}{2})$ $(sqrt(3)/2, -1/2)$ have $y$ $y$ values (or $sin$ $sin$ values) of $- \frac{1}{2}$ $-1/2$ .

The radian correspondents of these coordinates are $\frac{7 π}{6}$ $(7pi)/6$ and $\frac{11 π}{6}$ $(11pi)/6$ , and those are your two answers.

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