How do you solve $2 sin x - 1 = 0$ $2 sin x - 1 = 0$ over the interval 0 to 2pi?

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How do you solve $2 sin x - 1 = 0$ $2 sin x - 1 = 0$ over the interval 0 to 2pi?

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Answer 1 · 2018-04-14T10:27:38

$x = \frac{π}{6}, 5 \frac{π}{6}$ $x = pi/6 , 5pi/6$

Explanation:

1/ $2 sin (x) - 1 = 0$ $2sin(x) - 1 = 0$
2/ $2 sin (x) = 1$ $2sin(x) = 1$
3/ $sin (x) = \frac{1}{2}$ $sin(x) = 1/2$

4/ $x = \frac{π}{6}, 5 \frac{π}{6}$ $x = pi/6 , 5pi/6$

Answer 2 · 2018-04-14T10:28:22

$x = \frac{π}{6} or \frac{5 π}{6}$ $x=pi/6 or (5pi)/6$

Explanation:

$2 sin (x) - 1 = 0 ∣ + 1$ $2sin(x)-1=0|+1$
$2 sin (x) = 1 ∣ : 2$ $2sin(x)=1|:2$
$sin (x) = \frac{1}{2}$ $sin(x)=1/2$
$x = arcsin (\frac{1}{2}) = \frac{π}{6} or \frac{5 π}{6}$ $x=arcsin(1/2)=pi/6 or (5pi)/6$

Answer 3 · 2018-04-14T10:32:16

$x = \frac{π}{6}, \frac{5 π}{6}$ $x=pi/6,(5pi)/6$

Explanation:

$2 sin x - 1 = 0$ $2sinx-1=0$

$\Rightarrow sin x = \frac{1}{2}$ $rArrsinx=1/2$

$since sin x > 0 then x in first/second quadrant$ $"since "sinx>0" then x in first/second quadrant"$

$\Rightarrow x = {sin}^{- 1} (\frac{1}{2}) = \frac{π}{6} \leftarrow first quadrant$ $rArrx=sin^-1(1/2)=pi/6larrcolor(blue)"first quadrant"$

$or x = π - \frac{π}{6} = \frac{5 π}{6} \leftarrow second quadrant$ $"or "x=pi-pi/6=(5pi)/6larrcolor(blue)"second quadrant"$

$\Rightarrow x = \frac{π}{6}, \frac{5 π}{6} \to (0, 2 π)$ $rArrx=pi/6,(5pi)/6to(0,2pi)$

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How do you solve $2 sin x - 1 = 0$ $2 sin x - 1 = 0$ over the interval 0 to 2pi?

3 Answers

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