Question

2sin3x+sqrt3=0

Answer 1

start with
\[\sin(2x)=-\frac{\sqrt{3}}{2}\]

Answer 2

I got this but idk if it is right
2sin3x - sqrt3= 0
2sin3x = sqrt3
sin3x = (sqrt3) / 2
3x = pi/3, 2pi/3, 7pi/3, 8pi/3, 16pi/3, 17pi/3
x = pi/9, 2pi/9, 7pi/9, 8pi/9, 16pi/9, 17pi/9

Answer 3

one solution has \(2x=-\frac{\pi}{3}\) or if you prefer \(x=\frac{5\pi}{3}\) so \(2x=\frac{5\pi}{6}\) etc

Answer 4

oh damn i wrote \(2x\) when it should be \(3x\) sorry

Answer 5

no prob, but am i right

Answer 6

but you have a mistake in your solution

Answer 7

\[2\sin(x)+\sqrt{3}=0\iff 2\sin(x)=-\sqrt{3}\iff \sin(3x)=-\frac{\sqrt3}{2}\]

Answer 8

so it is not \(3x=\frac{\pi}{3}\) but rather \(3x=\frac{4\pi}{3},\frac{5\pi}{3}\) etc

Answer 9

hold on your posted problem has \(+\) but your solution has \(-\) in the equation, so i am not sure which is right

Answer 10

it is really + thats the problem i cant find the right answer for + and i only have 30 minutes to figure all this out