Solving for k

First, we simplify the expression for $k$. Let $\alpha = \frac{1}{2}\cos^{-1}\left(\frac{2}{3}\right)$. This means $\cos(2\alpha) = \frac{2}{3}$. Using the tangent half-angle identity related to cosine, $\cos(2\alpha) = \frac{1-\tan^2\alpha}{1+\tan^2\alpha}$, let $t = \tan\alpha$. We have:

$ \frac{1-t^2}{1+t^2} = \frac{2}{3} $

Solving for $t^2$: $3(1-t^2) = 2(1+t^2) \implies 3-3t^2 = 2+2t^2 \implies 5t^2 = 1 \implies t^2 = \frac{1}{5}$.

Since $2\alpha = \cos^{-1}(\frac{2}{3})$ is in $(0, \frac{\pi}{2})$, $\alpha$ is in $(0, \frac{\pi}{4})$, thus $\tan\alpha$ is positive.

$ \tan\alpha = \frac{1}{\sqrt{5}} $

Now, we use the tangent addition formula $\tan(\frac{\pi}{4}+\theta) = \frac{1+\tan\theta}{1-\tan\theta}$:

$ \tan\left(\frac{\pi}{4} + \alpha\right) = \frac{1+\tan\alpha}{1-\tan\alpha} = \frac{1 + \frac{1}{\sqrt{5}}}{1 - \frac{1}{\sqrt{5}}} = \frac{\sqrt{5}+1}{\sqrt{5}-1} $

Rationalizing the denominator:

$ \frac{(\sqrt{5}+1)(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{5+1+2\sqrt{5}}{5-1} = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2} $

Next, let $\beta = \frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)$. This means $\sin(2\beta) = \frac{2}{3}$. Using the tangent half-angle identity related to sine, $\sin(2\beta) = \frac{2\tan\beta}{1+\tan^2\beta}$, let $u = \tan\beta$. We have:

$ \frac{2u}{1+u^2} = \frac{2}{3} $

Solving for $u$: $3u = 1+u^2 \implies u^2 - 3u + 1 = 0$. Using the quadratic formula, $u = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.

Since $2\beta = \sin^{-1}(\frac{2}{3})$ is in $(0, \frac{\pi}{2})$, $\beta$ is in $(0, \frac{\pi}{4})$, thus $\tan\beta < 1$. Therefore:

$ \tan\beta = \frac{3-\sqrt{5}}{2} $

Now, substitute the calculated values back into the expression for $k$:

$ k = \tan\left(\frac{\pi}{4} + \alpha\right) + \tan\beta = \left(\frac{3+\sqrt{5}}{2}\right) + \left(\frac{3-\sqrt{5}}{2}\right) = \frac{3+\sqrt{5}+3-\sqrt{5}}{2} = \frac{6}{2} = 3 $

Solving the Inverse Trigonometric Equation

The equation is $\sin^{-1}(kx-1) = \sin^{-1}x - \cos^{-1}x$. Substitute $k=3$:

$ \sin^{-1}(3x-1) = \sin^{-1}x - \cos^{-1}x $

Using the identity $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$ (valid for $x \in [-1, 1]$):

$ \sin^{-1}(3x-1) = \sin^{-1}x - \left(\frac{\pi}{2} - \sin^{-1}x\right) $

$ \sin^{-1}(3x-1) = 2\sin^{-1}x - \frac{\pi}{2} $

Determine the domain constraints for $x$:

• For $\sin^{-1}x$ and $\cos^{-1}x$, we need $-1 \le x \le 1$.
• For $\sin^{-1}(3x-1)$, we need $-1 \le 3x-1 \le 1$, which implies $0 \le 3x \le 2$, or $0 \le x \le \frac{2}{3}$.
• The range of the right side, $2\sin^{-1}x - \frac{\pi}{2}$, must be within $[-\frac{\pi}{2}, \frac{\pi}{2}]$. This leads to $0 \le \sin^{-1}x \le \frac{\pi}{2}$, implying $0 \le x \le 1$.

Combining these constraints, the valid domain for $x$ is $0 \le x \le \frac{2}{3}$.

Finding Potential Solutions

Let $y = \sin^{-1}x$. Since $x \in [0, \frac{2}{3}]$, $y \in [0, \sin^{-1}(\frac{2}{3})]$. The equation becomes:

$ \sin^{-1}(3\sin y - 1) = 2y - \frac{\pi}{2} $

Apply the sine function to both sides:

$ 3\sin y - 1 = \sin\left(2y - \frac{\pi}{2}\right) $

Using the identity $\sin(\theta - \frac{\pi}{2}) = -\cos\theta$:

$ 3\sin y - 1 = -\cos(2y) $

Using the double angle identity $\cos(2y) = 1 - 2\sin^2y$:

$ 3\sin y - 1 = -(1 - 2\sin^2y) $

$ 3\sin y - 1 = -1 + 2\sin^2y $

$ 2\sin^2y - 3\sin y = 0 $

$ \sin y (2\sin y - 3) = 0 $

This gives $\sin y = 0$ or $\sin y = \frac{3}{2}$.

Since $x = \sin y$, the potential solutions are $x=0$ or $x=\frac{3}{2}$.

Verifying Solutions and Conclusion

We check these potential solutions against the domain $0 \le x \le \frac{2}{3}$:

• Check $x=0$: This value is within the domain. Substituting into the original equation: LHS: $\sin^{-1}(3(0)-1) = \sin^{-1}(-1) = -\frac{\pi}{2}$. RHS: $\sin^{-1}(0) - \cos^{-1}(0) = 0 - \frac{\pi}{2} = -\frac{\pi}{2}$. Since LHS = RHS, $x=0$ is a valid solution.
• Check $x=\frac{3}{2}$: This value is outside the domain $0 \le x \le \frac{2}{3}$. It is not a valid solution.

Therefore, there is exactly one solution to the equation.