The number of solutions of the equation $2 \operatorname{Cos}^{-1} x+\operatorname{Sin}^{-1} x=\frac{11…
The number of solutions of the equation $2 \operatorname{Cos}^{-1} x+\operatorname{Sin}^{-1} x=\frac{11 \pi}{6}$ is
$0$
$1$
$2$
$3$
Solution
The equation is $2 \operatorname{Cos}^{-1} x+\operatorname{Sin}^{-1} x=\frac{11 \pi}{6}$.
We use the identity $\operatorname{Cos}^{-1} x+\operatorname{Sin}^{-1} x=\frac{\pi}{2}$ for $x \in [-1, 1]$.
The equation can be rewritten as:
$(\operatorname{Cos}^{-1} x+\operatorname{Sin}^{-1} x) + \operatorname{Cos}^{-1} x = \frac{11 \pi}{6}$
$\frac{\pi}{2} + \operatorname{Cos}^{-1} x = \frac{11 \pi}{6}$
$\operatorname{Cos}^{-1} x = \frac{11 \pi}{6} - \frac{\pi}{2}$
$\operatorname{Cos}^{-1} x = \frac{11 \pi}{6} - \frac{3 \pi}{6}$
$\operatorname{Cos}^{-1} x = \frac{8 \pi}{6}$
$\operatorname{Cos}^{-1} x = \frac{4 \pi}{3}$
The range of the principal value of $\operatorname{Cos}^{-1} x$ is $[0, \pi]$.
Since $\frac{4 \pi}{3} > \pi$, there is no value of $x$ for which $\operatorname{Cos}^{-1} x = \frac{4 \pi}{3}$.
Therefore, the equation has $\mathbf{0}$ solutions.