Question

The equation $2{\cos }^{-1}x+{\sin }^{-1}x=\frac{11\mathrm{\pi }}{6}$ has.

• A. No solution
• B. Only one solution
• C. Two solutions
• D. Three solutions

Answer 1

The correct option is A

No solution

Explanation of correct answer :

Finding out the solution of the given equation :

Given, $2{\cos }^{-1}x+{\sin }^{-1}x=\frac{11\mathrm{\pi }}{6}$

Now solving,

$$\begin{gathered} \Rightarrow {\cos }^{-1}x+{\cos }^{-1}x+{\sin }^{-1}x=\frac{11\mathrm{\pi }}{6} \\ \Rightarrow {\cos }^{-1}x+\left({\cos }^{-1}x+{\sin }^{-1}x\right)=\frac{11\mathrm{\pi }}{6} \\ \Rightarrow {\cos }^{-1}x+\frac{\mathrm{\pi }}{2}=\frac{11\mathrm{\pi }}{6}\left({\cos }^{-1}x+{\sin }^{-1}x=\frac{\mathrm{\pi }}{2},\text{as per trigonometric identity}\right) \\ \Rightarrow {\cos }^{-1}x=\frac{11\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\cos }^{-1}x=\frac{4\mathrm{\pi }}{3} \end{gathered}$$

As we know, Range of ${\cos }^{-1}x\in \left[0,\mathrm{\pi }\right]$.

While solving the equation we got ${\cos }^{-1}x=\frac{4\mathrm{\pi }}{3}$ which does not lie in the range of ${\cos }^{-1}x$. So, it is not possible for any value.

Thus, no solution is possible for the given equation.

Hence, the correct answer is Option (A).