1. **State the problem:** Given $\tan x = -\frac{3}{4}$, find $\sin x$ and $\csc x$ for $x$ in the interval $0^\circ$ to $360^\circ$. 2. **Recall the identity:** $\tan x = \frac{\sin x}{\cos x}$. 3. **Set up a right triangle:** Let the opposite side be $-3$ and adjacent side be $4$ (since $\tan x$ is negative, the signs depend on the quadrant). 4. **Calculate the hypotenuse:** $$\text{hypotenuse} = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$ 5. **Determine the quadrant:** Since $\tan x = \frac{\sin x}{\cos x} = -\frac{3}{4}$ is negative, $x$ lies in either the second or fourth quadrant. 6. **Find $\sin x$ and $\cos x$ in each quadrant:** - In the second quadrant, $\sin x > 0$ and $\cos x < 0$. - In the fourth quadrant, $\sin x < 0$ and $\cos x > 0$. 7. **Calculate $\sin x$ and $\cos x$ for each quadrant:** - Second quadrant: $$\sin x = \frac{3}{5}, \quad \cos x = -\frac{4}{5}.$$ - Fourth quadrant: $$\sin x = -\frac{3}{5}, \quad \cos x = \frac{4}{5}.$$ 8. **Calculate $\csc x = \frac{1}{\sin x}$:** - Second quadrant: $$\csc x = \frac{1}{\frac{3}{5}} = \frac{5}{3}.$$ - Fourth quadrant: $$\csc x = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}.$$ **Final answers:** - For $x$ in the second quadrant: $$\sin x = \frac{3}{5}, \quad \csc x = \frac{5}{3}.$$ - For $x$ in the fourth quadrant: $$\sin x = -\frac{3}{5}, \quad \csc x = -\frac{5}{3}.$$