If k=(4+12^-123)+(12^-123), then the number of solutions of the equation ^-1(kx-1)=^-1x-^-…

Question

If $k=	an\left(\dfrac{\pi}{4}+\dfrac12\cos^{-1}\dfrac23ight)+	an\left(\dfrac12\sin^{-1}\dfrac23ight)$, then the number of solutions of the equation $\sin^{-1}(kx-1)=\sin^{-1}x-\cos^{-1}x$ is \_\_\_\_.

Answer: 1

Accepted Answer

Let $\theta=\cos^{-1}\dfrac23$, so $\sin^{-1}\dfrac23=\dfrac{\pi}{2}-\theta$. Then

$$k=\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)+\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)=\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}+\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}=\frac{2\left(1+\tan^2\frac{\theta}{2}\right)}{1-\tan^2\frac{\theta}{2}}=\frac{2}{\cos\theta}.$$

Since $\cos\theta=\dfrac23$, $k=\dfrac{2}{2/3}=3$.

Now solve $\sin^{-1}(3x-1)=\sin^{-1}x-\cos^{-1}x=2\sin^{-1}x-\dfrac{\pi}{2}$.

Domain: $3x-1\in[-1,1]\Rightarrow x\in\left[0,\dfrac23\right]$.

Taking sine of both sides:
$$3x-1=\sin\left(2\sin^{-1}x-\frac{\pi}{2}\right)=-\cos(2\sin^{-1}x)=-(1-2x^2)=2x^2-1.$$

So $3x=2x^2\Rightarrow x(2x-3)=0\Rightarrow x=0\ \text{or}\ x=\dfrac32$.

Only $x=0$ lies in $\left[0,\dfrac23\right]$. Hence the number of solutions is 1.

Solutions · 1