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: Use the standard identity.
For any valid $x$, $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$.

Step 2: Split the given equation.
Write $4\sin^{-1}x+6\cos^{-1}x$ as $4(\sin^{-1}x+\cos^{-1}x)+2\cos^{-1}x$. So $4\cdot\frac{\pi}{2}+2\cos^{-1}x=3\pi$.

Step 3: Solve for the inverse cosine.
$2\pi+2\cos^{-1}x=3\pi$, so $2\cos^{-1}x=\pi$, giving $\cos^{-1}x=\frac{\pi}{2}$.

Step 4: Find $x$.
$x=\cos\frac{\pi}{2}=0$. \[ \boxed{x=0} \]