## Task B-3.6.

For which values of the real parameter $a$ does the equation $a \sin x - \tan x = 1$ have a solution in the interval $\left(0, \frac{\pi}{2}\right)$?
## Solution.

Let's write the given equation in the form $a \sin x=\operatorname{tg} x+1$.

Notice that the number $a$ must be positive and different from 0 because $x$ is in the interval $\left\langle 0, \frac{\pi}{2}\right\rangle$ where both $\sin x$ and $\operatorname{tg} x$ are positive.

$$
\begin{gathered}
a \sin x=\frac{\sin x+\cos x}{\cos x} \\
a \sin x \cos x=\sin x+\cos x \\
a^{2} \sin ^{2} x \cos ^{2} x=(\sin x+\cos x)^{2} \\
\frac{1}{4} a^{2} \sin ^{2} 2 x=1+2 \sin x \cos x \\
a^{2} \sin ^{2} 2 x=4+4 \sin 2 x
\end{gathered}
$$

With the substitution $y=\sin 2 x$, we get the quadratic equation $a^{2} y^{2}-4 y-4=0$ whose solutions are

$$
y_{1,2}=\frac{4 \pm \sqrt{16+16 a^{2}}}{2 a^{2}}=\frac{2 \pm 2 \sqrt{1+a^{2}}}{a^{2}}
$$

Since $x \in\left\langle 0, \frac{\pi}{2}\right\rangle$, it follows that $2 x \in\langle 0, \pi\rangle$ and $\sin 2 x>0$.

Therefore, $y=\frac{2+2 \sqrt{1+a^{2}}}{a^{2}}$, with the condition $y \leqslant 1$, which is equivalent to

$$
2 \sqrt{1+a^{2}} \leqslant a^{2}-2
$$

The left side of this inequality is positive, so it must be $a^{2}-2>0$, i.e., $|a|>\sqrt{2}$, or, since $a>0$, $a>\sqrt{2}$.

(*) 1 point

With this condition, after squaring, we get

$$
\begin{array}{ll}
4+4 a^{2} \leqslant a^{4}-4 a^{2}+4 & 1 \text { point } \\
a^{4}-8 a^{2} \geqslant 0 & 1 \text { point } \\
a^{2} \geqslant 8 \quad|a| \geqslant 2 \sqrt{2} & 1 \text { point }
\end{array}
$$

From $\left({ }^{*}\right)$, $a>\sqrt{2}$, so it follows that $a \geqslant 2 \sqrt{2}$, i.e., $a \in[2 \sqrt{2}, \infty\rangle$.

Note: If the student did not mention the initial condition $a>0$ and the condition $\left(^{*}\right)$ that $a^{2}>2$, or $|a|>\sqrt{2}$, during the solution of the inequality, 2 points should be deducted, regardless of the correct solution.