Solving with an inverse

[study24/7]

I'm unsure if I received the correct answer. I am supposed to solve with an inverse so I got (3sin(J*sqrt(2)) /2. I used the inverse of sin to solve it

 

[Deleted member 4993]

View attachment 20485

I'm unsure if I received the correct answer. I am supposed to solve with an inverse so I got (3sin(J*sqrt(2)) /2. I used the inverse of sin to solve it

That is incorrect.

Please show work, step-by-step, so that we can correct your mistake. If I were to do this problem, I would start with:

\(\displaystyle \sqrt{2} * \sin(\frac{2}{3}*\theta) = J\)

\(\displaystyle \sin(\frac{2}{3}*\theta) = \frac {J \sqrt{2}}{2}\)

\(\displaystyle \frac{2}{3}*\theta = \sin^{-1}\left(\frac {J \sqrt{2}}{2}\right)\)

continue.....