## Trigonometry 7th Edition

$(a)$ $\displaystyle \theta=\frac{\pi}{3}+2k\pi$ or $\displaystyle \theta=\frac{2\pi}{3}+2k\pi$ $(b)$ $\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{2\pi}{3}.$
The first task is to isolate the trigonometric function on one side: Add $(\sqrt{3}-2\sin t)$ to both sides... $4\sin t-2\sin t=\sqrt{3}\qquad$ ... simplify $2\sin t=\sqrt{3}\quad$ ... divide with $2$ $\displaystyle \sin t=\frac{\sqrt{3}}{2}$ Now, we find a reference angle. From the table of characteristic angles, we know that $\displaystyle \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}.$ Next, we know that sine is positive in quadrants I and II, so angles (in radians) within the interval $0\leq t \lt 2\pi$ that satisfy the equation are $\displaystyle \frac{\pi}{3}$ and $\pi-\displaystyle \frac{\pi}{3}=\frac{2\pi}{3}$ Finally, to each individual solution, add multiples of $2\pi$ to cover all solutions: $(a)$ $\displaystyle \theta=\frac{\pi}{3}+2k\pi$ or $\displaystyle \theta=\frac{2\pi}{3}+2k\pi$ $(b)$ The solutions within the interval $0\leq t \lt 2\pi:$ $\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{2\pi}{3}.$