Answer
$\frac{4-3\sqrt 2}{4}$
Work Step by Step
Find $dy/d\theta$:
$\frac{dy}{d\theta}=\frac{d}{d\theta}(r\sin\theta)=\frac{d}{d\theta}((2+\sin3\theta)\sin\theta)$
$=\frac{d}{d\theta}(2+\sin 3\theta)\sin\theta+(2+\sin3\theta)\frac{d}{d\theta}(\sin\theta)$
$=3\cos3\theta\sin\theta+(2+\sin3\theta)\cos\theta$
$=3\cos3\theta\sin\theta+\sin3\theta\cos\theta+2\cos\theta$
Find $dx/d\theta$:
$\frac{dx}{d\theta}(r\cos\theta)=\frac{d}{d\theta}((2+\sin3\theta)\cos\theta)$
$=\frac{d}{d\theta}(2+\sin3\theta)\cdot \cos\theta+(2+\sin3\theta)\frac{d}{d\theta}\cos\theta)$
$=3\cos3\theta\cos\theta+(2+\sin3\theta)(-\sin\theta)$
$=3\cos3\theta\cos\theta-\sin3\theta\sin\theta-2\sin\theta$
Find the slope of the tangent line at $\theta=\frac{\pi}{4}$
$m=\frac{dy}{dx}|_{\theta=\frac{\pi}{4}}=\frac{dy/d\theta}{dx/d\theta}|_{\theta=\frac{\pi}{4}}$
$=\frac{3\cos\frac{3\pi}{4}\sin\frac{\pi}{4}+\sin\frac{3\pi}{4}\cos\frac{\pi}{4}+2\cos\frac{\pi}{4}}{3\cos\frac{3\pi}{4}\cos\frac{\pi}{4}-\sin\frac{3\pi}{4}\sin\frac{\pi}{4}-2\sin\frac{\pi}{4}}$
$=\frac{3\cdot (-\frac{\sqrt{2}}{2})\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2}}{2}+2\cdot\frac{\sqrt{2}}{2}}{3\cdot(-\frac{\sqrt{2}}{2})\cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2}-2\frac{\sqrt{2}}{2}}$
$=\frac{-\frac{3}{2}+\frac{1}{2}+\sqrt{2}}{-\frac{3}{2}-\frac{1}{2}-\sqrt{2}}$
$=\frac{1-\sqrt{2}}{2+\sqrt{2}}$
Thus, the slope is $\frac{1-\sqrt{2}}{2+\sqrt{2}}=\frac{4-3\sqrt 2}{4}$.
