Question

The period of the sine function is 2pi. In the interval [0, 2pi) there are two angles .69x for which sin(.69x)=-4/5. please help?

Answer 1

\[\sin(0.77x)=-\frac{1}{2}\\
0.77x=\arcsin\left(-\frac{1}{2}\right)\]
In the interval \([0,2\pi)\), sinx gives you -1/2 for \(x=\dfrac{7\pi}{6},\dfrac{11\pi}{6}\approx 3.665,~5.760\).
So you have the equations
\[\begin{matrix}0.77x=\dfrac{7\pi}{6}&&&0.77x=\dfrac{11\pi}{6}\end{matrix}\]
Or, approximately
\[\begin{matrix}0.77x=3.665&&&0.77x=5.760\end{matrix}\]

Answer 2

thank you so much!

Answer 3

what about sin(.69x)=-4/5 with the same interval

Answer 4

You're welcome!

For your second question: I don't know the value of \(\sin^{-1}\left(-\dfrac{4}{5}\right)\) right off the top of my head, so I'd have to use a calculator...

\(\sin x=-\dfrac{4}{5}\) when \(x\approx 4.069,~5.356\) for \(0\le x<2\pi\), so you have
\[\begin{matrix}
0.69x=4.069&&&0.69x=5.356
\end{matrix}\]

Answer 5

I actually figured out a super easy way to do it! If you graph it on a calculator making y1=sin(.69x) and y2=-4/5, just look for the intercepts and it tells you the answers haha