If [ x ] denotes the greatest integer ≤ x , then the system of linear equations [ sin θ ] x + [ − cos θ ] y = 0 [ cot θ ] x + y = 0 has

1. Greatest-Integer (Floor) Function refresher

For any real number x, the symbol $[x]$ means the largest integer ≤ x.

• If x = 0.7, then $[x] = 0$.
• If x = −0.2, then $[x] = −1$.

Because $\sin\theta$ and $\cos\theta$ live inside [−1, 1], the possible floor values are −1, 0, or 1 (with 1 or −1 happening only at very special angle points).

2. Writing the determinant

For a 2×2 system

What does the question say?

Imagine you have two see-saws (the two equations). The long sticks x and y will balance the see-saws only when the weights written in square-brackets are fixed by the angle theta (θ).

[sinθ]·x + [−cosθ]·y = 0
[cotθ]·x +      y     = 0

Here [ ] just means round down to the nearest whole number (the greatest integer ≤ the number).

When do we get more than the boring (x = 0, y = 0) balance?

If the two see-saws are really the same (or one is a multiple of the other), there will be many ways to balance — infinitely many (non-zero) (x, y). Mathematically, that happens when the little square-bracket numbers make the two equations dependent (their determinant becomes 0).

What angles give that situation?

Because sinθ and cosθ always stay between −1 and 1, their rounded-down values can only be −1, 0, or 1. A bit of checking shows that the only realistic way to make the determinant 0 (and hence get infinitely many solutions) is when

• $[\sin\theta] = 0$ (so $0 \le \sin\theta < 1$)
• $[\!\;-\cos\theta] = 0$ (so $0 \le -\cos\theta < 1 \;\Rightarrow\; \cos\theta \le 0$)

That puts θ squarely in the second quadrant, i.e. $\tfrac{\pi}{2} < \theta < \pi$.

Any angle there gives us infinitely many non-trivial solutions. Everywhere else the two see-saws are truly different, so the only balance is the boring (0, 0).

Step 1: Write the determinant

D = \begin{vmatrix}[\sin\theta] & [\!\;-\cos\theta]\\[\cot\theta] & 1\end{vmatrix} = [\sin\theta] - [\!\;-\cos\theta]\,[\cot\theta].

Step 2: Look at possible floor values

Because $\sin\theta,\cos\theta \in [-1,1]$, $[\sin\theta],\;[\!\;-\cos\theta] \in \{-1,\;0,\;1\}.$

Step 3: Force $D=0$ (for non-trivial solutions)

The simplest way is to make $[\!\;-\cos\theta]=0$. Then $D=[\sin\theta]$; forcing that also to 0 gives $D=0$.

So we require $[\sin\theta]=0 \quad\text{and}\quad [\!\;-\cos\theta]=0.$

Step 4: Translate each floor condition

1. $[\sin\theta]=0 \;\Rightarrow\; 0 \le \sin\theta < 1$.
2. $[\!\;-\cos\theta]=0 \;\Rightarrow\; 0 \le -\cos\theta < 1 \;\Rightarrow\; \cos\theta \le 0.$

Those two together give the interval $\boxed{\dfrac{\pi}{2} < \theta < \pi}.$

Step 5: State the outcome

For every $\theta$ in that interval, the determinant is 0 and the two equations are dependent, so the system has infinitely many non-trivial solutions (x, y). For all other θ, the determinant is non-zero and the only solution is the trivial $(0,0)$.