1. Given \( \tan x = \frac{3}{4} \), we need to find \( \sin x \) and \( \csc x \) for \( x \) in the interval \( 0^\circ \) to \( 360^\circ \). 2. Recall that \( \tan x = \frac{\sin x}{\cos x} \). So, if \( \tan x = \frac{3}{4} \), then \( \sin x = 3k \) and \( \cos x = 4k \) for some \( k \). 3. Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), substitute: $$ (3k)^2 + (4k)^2 = 1 $$ $$ 9k^2 + 16k^2 = 1 $$ $$ 25k^2 = 1 $$ $$ k^2 = \frac{1}{25} $$ $$ k = \pm \frac{1}{5} $$ 4. Therefore, \( \sin x = 3k = \pm \frac{3}{5} \) and \( \cos x = 4k = \pm \frac{4}{5} \). 5. Since \( \tan x = \frac{3}{4} > 0 \), \( x \) lies in the first or third quadrant where tangent is positive. 6. In the first quadrant, both sine and cosine are positive: \( \sin x = \frac{3}{5} \), \( \cos x = \frac{4}{5} \). 7. In the third quadrant, both sine and cosine are negative: \( \sin x = -\frac{3}{5} \), \( \cos x = -\frac{4}{5} \). 8. The cosecant is the reciprocal of sine: \( \csc x = \frac{1}{\sin x} \). 9. So, for the first quadrant: \( \csc x = \frac{1}{\frac{3}{5}} = \frac{5}{3} \). 10. For the third quadrant: \( \csc x = \frac{1}{-\frac{3}{5}} = -\frac{5}{3} \). **Final answers:** - \( \sin x = \frac{3}{5} \) or \( -\frac{3}{5} \) - \( \csc x = \frac{5}{3} \) or \( -\frac{5}{3} \)