Trigonometric equation over unit circle

Question

Let us consider the equations $$ {p_0}^2 + {q_0}^2 =1 \qquad\text{and}\qquad {-a} p_0\sin s + b q_0\cos s = 0 $$ How do i solve for $p_0, q_0$?

Answer 1

Let $s=0$ so that we have $bq_0=0$, and by letting $s=\pi/2$ we get $-ap_0=0$. Then either $b=0$ or $q_0=0$. If the latter, then $p_0\in\{-1,1\}$ by the first equation. Similarly, if $p_0=0$ then $q_0\in\{-1,1\}$. So the solutions $(p_0,q_0)$ must be among $(-1,0)$, $(1,0)$ if $a=0$ and $(0,-1)$, $(0,1)$ if $b=0$. It is routine to verify that these are indeed solutions in the given cases.

Answer 2

The first equation is the unit circle and the second is a line passing through the origin. You can solve the two equations to find the points of intersections.

You can parametrize the unit circle $ x^2 + y^2 = 1 $ as $(x, y) = (\cos \theta, \sin \theta) $ for $\theta \in [0, 2 \pi)$.

Substituting in the second equation, you have $ \tan \theta = \frac{a}{b} \tan s$, or $\theta = \arctan(\frac{a}{b} \tan s)$ or $\pi +\arctan(\frac{a}{b} \tan s)$.

Answer 3

Let us rewrite the system in the simple form

$$\begin{cases}p^2+q^2=1,\\up=vq.\end{cases}$$

Then

$$v^2p^2+v^2q^2=v^2$$ is

$$(u^2+v^2)p^2=v^2.$$

The rest is yours.