$\displaystyle (cos \theta)x - (sin \theta)y=2$
$\displaystyle (sin\theta)x + (cos \theta)y=1$
In the range $\displaystyle 0 \leq \theta < 2 \pi$ is it solvable? Or are there values of theta when the equations are not solvable?
$\displaystyle (cos \theta)x - (sin \theta)y=2$
$\displaystyle (sin\theta)x + (cos \theta)y=1$
In the range $\displaystyle 0 \leq \theta < 2 \pi$ is it solvable? Or are there values of theta when the equations are not solvable?
Assuming you mean to solve for x and y, just treat $\displaystyle sin(\theta)$ and $\displaystyle cos(\theta)$ as constants. In particular, you can multiply the first equation by $\displaystyle cos(\theta)$, the second equation by $\displaystyle sin(\theta)$, and add the equations to eliminate y. A system of equations would be "non-solvable" if and only if the resulting coefficient of x were 0. But a rather nice thing happens here!