Question

Find the set of values of 'a' for which the equation $2 \cos^{-1}x = a + a^2 (\cos^{-1}x)^{-1}$ posses a solution.

Answer 1

[-2\pi, \pi]

Answer 2

Let $y = \cos^{-1}x$. The range of $\cos^{-1}x$ is $[0, \pi]$, so $y \in [0, \pi]$. The given equation transforms to $2y = a + \frac{a^2}{y}$.

Case 1: $y = 0$. If $y=0$, then $\cos^{-1}x = 0$, which means $x=1$. Substituting $y=0$ into $2y = a + \frac{a^2}{y}$ yields $0 = a + \frac{a^2}{0}$. If $a \neq 0$, $\frac{a^2}{0}$ is undefined, so this is impossible. If $a = 0$, the original equation becomes $2 \cos^{-1}x = 0$, which implies $\cos^{-1}x = 0$, so $x=1$. This is a valid solution. Thus, $a=0$ is a possible value.

Case 2: $y \in (0, \pi]$. Since $y \neq 0$, we can multiply by $y$: $2y^2 = ay + a^2$. Rearranging gives a quadratic equation in $y$: $2y^2 - ay - a^2 = 0$. Using the quadratic formula, $y = \frac{a \pm \sqrt{(-a)^2 - 4(2)(-a^2)}}{4} = \frac{a \pm \sqrt{a^2 + 8a^2}}{4} = \frac{a \pm \sqrt{9a^2}}{4} = \frac{a \pm 3|a|}{4}$. The two potential roots are $y_1 = \frac{a + 3|a|}{4}$ and $y_2 = \frac{a - 3|a|}{4}$. We need at least one root to be in $(0, \pi]$.

Subcase 2.1: $a > 0$. Then $|a| = a$. $y_1 = \frac{a + 3a}{4} = \frac{4a}{4} = a$. For $y_1$ to be in $(0, \pi]$, we need $0 < a \le \pi$. $y_2 = \frac{a - 3a}{4} = \frac{-2a}{4} = -\frac{a}{2}$. Since $a>0$, $y_2 < 0$, so it's not in $(0, \pi]$. Thus, for $a > 0$, the condition is $a \in (0, \pi]$.

Subcase 2.2: $a < 0$. Then $|a| = -a$. $y_1 = \frac{a + 3(-a)}{4} = \frac{-2a}{4} = -\frac{a}{2}$. For $y_1$ to be in $(0, \pi]$, we need $0 < -\frac{a}{2} \le \pi$. This implies $0 < -a \le 2\pi$, so $-2\pi \le a < 0$. $y_2 = \frac{a - 3(-a)}{4} = \frac{4a}{4} = a$. Since $a<0$, $y_2$ is not in $(0, \pi]$. Thus, for $a < 0$, the condition is $a \in [-2\pi, 0)$.

Combining the results for $y \in (0, \pi]$: $a \in [-2\pi, 0) \cup (0, \pi]$. Including the result from Case 1 ($a=0$), the complete set of values for 'a' is $[-2\pi, 0) \cup (0, \pi] \cup \{0\} = [-2\pi, \pi]$.