## Precalculus (6th Edition) Blitzer

If $\sin 0.9695\approx 0.8246$, then the solutions of $\sin x=-0.8246$, $0\le x\le 2\pi$, are given by $x\approx \underline{2\pi }+0.9695$ and $x\approx \underline{2\pi }-0.9695$.
It is given in the problem that the values lie between $0\le x\le 2\pi$. And the values of the sine function are negative in the third and fourth quadrant. When 0.9695 is added to $2\pi$ the value remains negative and lies in the fourth quadrant and if $2\pi$ is subtracted from 0.9695, the value again remains negative. Thus, $x\approx 2\pi +0.9695$ and $x=2\pi -0.9695$.