From the given condition: \[ 2\sin^{-1} x - 3\cos^{-1} x = 4x. \] Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, we can express $\cos^{-1} x$ as: \[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x. \] Substitute this into the original equation: \[ 2\sin^{-1} x - 3\left(\frac{\pi}{2} - \sin^{-1} x\right) = 4x \] Simplify: \[ 2\sin^{-1} x - \frac{3\pi}{2} + 3\sin^{-1} x = 4x \] Combine like terms: \[ 5\sin^{-1} x - \frac{3\pi}{2} = 4x \] Solve for $5\sin^{-1} x$: \[ 5\sin^{-1} x = 4x + \frac{3\pi}{2} \] Now, find $2\sin^{-1} x + 3\cos^{-1} x$: \[ 2\sin^{-1} x + 3\left(\frac{\pi}{2} - \sin^{-1} x\right) = 2\sin^{-1} x + \frac{3\pi}{2} - 3\sin^{-1} x = \frac{3\pi}{2} - \sin^{-1} x \] From the earlier equation: \[ \sin^{-1} x = \frac{4x + \frac{3\pi}{2}}{5} \] Substitute back: \[ 2\sin^{-1} x + 3\cos^{-1} x = \frac{3\pi}{2} - \frac{4x + \frac{3\pi}{2}}{5} = \frac{15\pi}{10} - \frac{4x + 15\pi}{10} = \frac{6\pi - 4x}{5} \] Given $2\sin^{-1} x - 3\cos^{-1} x = 4x$, we can solve for $x$ to find the exact value.
However, the simplified expression for $2\sin^{-1} x + 3\cos^{-1} x$ is $\frac{6\pi - 4}{5}$ when evaluated at the specific solution.
Hence, the correct answer is $\frac{6\pi - 4}{5}$.
1. Understand the problem:
Given 2sin⁻¹x - 3cos⁻¹x = 4 for x ∈ [-1,1], we need to find the value of 2sin⁻¹x + 3cos⁻¹x.
2. Use the identity sin⁻¹x + cos⁻¹x = π/2:
Let sin⁻¹x = α and cos⁻¹x = β. We know α + β = π/2.
3. Set up equations:
From given: 2α - 3β = 4
From identity: α + β = π/2
4. Solve the system:
From identity: α = π/2 - β
Substitute into first equation: 2(π/2 - β) - 3β = 4 ⇒ π - 2β - 3β = 4 ⇒ π - 5β = 4 ⇒ β = (π - 4)/5
Then α = π/2 - (π - 4)/5 = (5π - 2π + 8)/10 = (3π + 8)/10
5. Find required expression:
2α + 3β = 2(3π + 8)/10 + 3(π - 4)/5 = (6π + 16)/10 + (3π - 12)/5 = (6π + 16 + 6π - 24)/10 = (12π - 8)/10 = (6π - 4)/5
Correct Answer: (B) \(\frac{6\pi-4}{5}\)