## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$r=\dfrac{\sqrt 6}{2}\quad \quad \text{or}\quad \quad r^2=\dfrac{3}{2}$
Recall that $x=r\cos\theta$ and $y=r\sin\theta$. Substitute these in the given equation $2x^{2}+2y^{2}=3$ to obtain: $2(r\cos\theta)^{2}+2(r\sin\theta)^{2}=3$ $2r^2\cos^2\theta+2r^2\sin^2\theta=3$ Factor out $2r^2$ to obtain: $2r^{2}(\cos^{2}\theta+\sin^{2}\theta)=3$ Since $\cos^{2}\theta+\sin^{2}\theta=1$, then the equation above becomes: $2r^{2}=3$ $r^{2}=\frac{3}{2} \implies r=\frac{\sqrt 3}{\sqrt 2}=\frac{\sqrt 3}{\sqrt2}\cdot \frac{\sqrt2}{\sqrt2}=\frac{\sqrt 6}{2}$ Thus, the polar form of the given equation is: $$r=\dfrac{\sqrt 6}{2}\quad \quad \text{or}\quad \quad r^2=\dfrac{3}{2}$$