Question

solve on the interval [0,2Pi): 2sintheta + square root of 3 = 0

Answer 1

\[2\sin \theta = - \sqrt{3}\]
\[\sin \theta = - \frac{ \sqrt{3} }{ 2 }\]
Much easier to solve now

Answer 2

take sin^-1 of - rad (3) / 2
There are 2 ans
1. 4pi/3, 5pi/3

Answer 3

it's 4 pi/3, the 1. is just there c:

Answer 4

whats that then

Answer 5

i mean theta

Answer 6

\[\frac{ 4\pi }{ 3 }\] and \[\frac{ 5 \pi }{ 3 }\] and the steps to do it is mentioned above

Answer 7

thanks very much

Answer 8

@abdollar19 welcome
@Bebong thanks c:

Answer 9

where u from