015 11: Problem 17
(1 point)
Solve the following equation in the interval $[0,2 \pi]$.
Note: Give the answer as a multiple of $\pi$. Do not use decimal numbers. The answer should be fraction or an integer. Note that $\pi$ is already included in the answer so you just have to enter th appropriate multiple. E.g. if the answer is $\pi / 2$ you should enter $1 / 2$. If there is more than one an enter them separated by commas.
\[
\begin{array}{l}
2(\sin (t))^{2}-\sin (t)-1=0 \\
\mathrm{t}=\square \pi
\end{array}
\]

Step 1 ：Define the given equation as \(2\sin(t)^2 - \sin(t) - 1 = 0\).

Step 2 ：Solve the equation for \(t\). The solutions are \(-\pi/6\), \(\pi/2\), and \(7\pi/6\).

Step 3 ：However, the question asks for the solutions in the interval \([0, 2\pi]\). The solution \(-\pi/6\) is not in this interval.

Step 4 ：Add \(2\pi\) to the solution \(-\pi/6\) to get a solution in the interval \([0, 2\pi]\). The solution \(-\pi/6 + 2\pi\) is equal to \(11\pi/6\).

Step 5 ：Therefore, the solutions for \(t\) in the interval \([0, 2\pi]\) are \(\pi/2\), \(7\pi/6\), and \(11\pi/6\).

Step 6 ：Final Answer: \(\boxed{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}}\)