Question

If $4 \sin^{-1} x + \cos^{-1} x = \pi$ then $x =$

• A. $\frac{\sqrt{3}}{2}$
• B. 0
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Hint:

Use identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The given algebraic equation features mixed inverse trigonometric functions. The most effective strategy is to eliminate one of the functions to create an equation in a single variable. We achieve this by using the fundamental complementary angle identity linking inverse sine and inverse cosine. Step 2: Key Formula or Approach:
Use the core identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all $x \in [-1, 1]$. Step 3: Detailed Explanation:
The given equation is: \[ 4 \sin^{-1} x + \cos^{-1} x = \pi \] We can decompose the term $4 \sin^{-1} x$ to expose a part that matches our identity: \[ 3 \sin^{-1} x + (\sin^{-1} x + \cos^{-1} x) = \pi \] Substitute the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ into the parentheses: \[ 3 \sin^{-1} x + \frac{\pi}{2} = \pi \] Now, isolate the term with $x$: \[ 3 \sin^{-1} x = \pi - \frac{\pi}{2} \] \[ 3 \sin^{-1} x = \frac{\pi}{2} \] Divide both sides by 3: \[ \sin^{-1} x = \frac{\pi}{6} \] To solve for $x$, apply the sine function to both sides: \[ x = \sin\left(\frac{\pi}{6}\right) \] From standard trigonometric values, we know that $\sin(30^\circ) = \sin(\pi/6) = \frac{1}{2}$. Therefore: \[ x = \frac{1}{2} \] Step 4: Final Answer:
The value of $x$ is $\frac{1}{2}$.