Question

If $4 \sin ^{-1} x+6 \cos ^{-1} x=3 \pi$, where $-1 \leq x \leq 1$, then $x =$

• A. 12
• B. 12
• C. -12
• D. 0

Hint:

Whenever you see a mixed combination of $\sin^{-1}x$ and $\cos^{-1}x$, always break down the larger coefficient to form a clean group of $(\sin^{-1}x + \cos^{-1}x)$. This trick instantly collapses complex expressions into simple values!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an equation containing inverse sine and inverse cosine functions. We need to solve for the specific value of $x$ that satisfies this relationship.

Step 2: Key Formula or Approach:
We use the fundamental inverse trigonometric identity that links complementary functions: $$ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} $$

Step 3: Detailed Explanation:
Let's rewrite the given expression by splitting the $6\cos^{-1} x$ term to group a complementary component: $$ 4\sin^{-1} x + 4\cos^{-1} x + 2\cos^{-1} x = 3\pi $$ Factor out the common coefficient 4 from the first two terms: $$ 4(\sin^{-1} x + \cos^{-1} x) + 2\cos^{-1} x = 3\pi $$ Substitute the standard identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ into our equation: $$ 4\left(\frac{\pi}{2}\right) + 2\cos^{-1} x = 3\pi $$ $$ 2\pi + 2\cos^{-1} x = 3\pi $$ Isolate the inverse trigonometric term by subtracting $2\pi$ from both sides: $$ 2\cos^{-1} x = 3\pi - 2\pi = \pi $$ $$ \cos^{-1} x = \frac{\pi}{2} $$ To isolate $x$, take the cosine on both sides: $$ x = \cos\left(\frac{\pi}{2}\right) $$ Since the cosine of $90^\circ$ is exactly zero: $$ x = 0 $$

Step 4: Final Answer:
The value of $x$ is 0, which corresponds to option (D).