## Trigonometry 7th Edition

a) $48.6^{\circ} + 360^{\circ}k$ and $131.4^{\circ} + 360^{\circ}k$, for all integers $k$. b) $\theta = 48.6˚, 131.4˚$
b) $4sin\theta - 3 = 0$ $4sin\theta = 3$ $sin\theta = \frac{3}{4}$ by GDC / calculator $\theta = sin^{-1}(0.75)$ $\theta = 48.59˚$ $\theta = 48.6˚$ But there are two answers, since $sin\theta$ is positive in both quadrant I and II. $= 180 - 48.6$ $= 131.4˚$ Therefore, $\theta = 48.6˚, 131.4˚$ a) Take the two angles and add $360^{\circ}k$for all integers $k$, to find all degree solutions. $48.6^{\circ} + 360^{\circ}k$ and $131.4^{\circ} + 360^{\circ}k$, for all integers $k$.