Question

please help me i need to find θ - the question is sinθ=negative square root 3 /2 , θ ~[-π ,π] an explanation would be appreciated :)

Answer 1

great -.- a smarter answer perhaps?

Answer 2

So you have:

\[\sinθ = -\frac{ \sqrt{3} }{ 2 }\]

to find θ take the arcsin of both sides.
\[\sin^{-1} (\sinθ) = \sin^{-1} -\frac{ \sqrt{3} }{ 2}\]
You'd get:
θ = -60 degrees or in

radians:

θ = \[-\frac{ π }{ 3 }\]

Think this is what you were asking for.

Answer 3

thankyou so much :)

Answer 4

No problem :)