How do you solve $2 sin (\frac{3 x}{2}) + \sqrt{3} = 0$ $2sin((3x)/2)+sqrt3=0$ ?

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How do you solve $2 sin (\frac{3 x}{2}) + \sqrt{3} = 0$ $2sin((3x)/2)+sqrt3=0$ ?

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Answer 1 · 2016-08-07T02:42:03

$\frac{8 π}{9} + 2 k π$ $(8pi)/9 + 2kpi$
$\frac{10 π}{9} + 2 k π$ $(10pi)/9 + 2kpi$

Explanation:

$2 sin (\frac{3 x}{2}) + \sqrt{3} = 0$ $2sin ((3x)/2) + sqrt3 = 0$
$sin (\frac{3 x}{2}) = - \frac{\sqrt{3}}{2}$ $sin ((3x)/2) = - sqrt3/2$
Trig table of special arcs, and unit circle give 2 arcs (3x)/2:
arc $\frac{3 x}{2} = - \frac{2 π}{3}$ $(3x)/2 = - (2pi)/3$ , and arc $\frac{3 x}{2} = π - (\frac{- 2 π}{3}) = \frac{5 π}{3}$ $(3x)/2 = pi - ((-2pi)/3) = (5pi)/3$
a. Arc $\frac{- 2 π}{3}$ $(-2pi)/3$ --> is the same as arc $\frac{4 π}{3}$ $(4pi)/3$ (co-terminal)
$\frac{3 x}{2} = \frac{4 π}{3}$ $(3x)/2 = (4pi)/3$ --> $x = (4 \frac{π}{3}) (\frac{2}{3}) = \frac{8 π}{9}$ $x = (4pi/3)(2/3) = (8pi)/9$

b. $\frac{3 x}{2} = \frac{5 π}{3}$ $(3x)/2 = (5pi)/3$ --> $x = (5 \frac{π}{3}) (\frac{2}{3}) = \frac{10 π}{9}$ $x = (5pi/3)(2/3) = (10pi)/9$
General answers:
$x = \frac{8 π}{9} + 2 k π$ $x = (8pi)/9 + 2kpi$
$x = \frac{10 π}{9} + 2 k π$ $x = (10pi)/9 + 2kpi$

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How do you solve $2 sin (\frac{3 x}{2}) + \sqrt{3} = 0$ $2sin((3x)/2)+sqrt3=0$ ?

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Explanation:

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How do you solve $2 sin (\frac{3 x}{2}) + \sqrt{3} = 0$ $2sin((3x)/2)+sqrt3=0$ ?