If k= (( )/(4)+(1)/(2) ^-1((2)/(3)))+ ((1)/(2) ^-1((2)/(3))), then the number of solutions…

Question

If $k=	an (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}(\frac{2}{3}))+	an (\frac{1}{2}{\sin }^{-1}(\frac{2}{3}))$, then

the number of solutions of the equation ${\sin }^{-1}(kx-1)={\sin }^{-1}x-{\cos }^{-1}x$ is $____$.

Answer: 1

Accepted Answer

Given :

$k=	an (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}(\frac{2}{3}))+	an (\frac{1}{2}{\sin }^{-1}(\frac{2}{3}))$

Let ${\cos }^{-1}\frac{2}{3}=	heta $, so ${\sin }^{-1}\frac{2}{3}=\frac{\pi }{2}-	heta $

$k=	an (\frac{\pi }{4}+\frac{	heta }{2})+	an (\frac{\pi }{4}-\frac{	heta }{2})$

$\Rightarrow$ $k=\frac{1+	an \frac{	heta }{2}}{1-	an \frac{	heta }{2}}+\frac{1-	an \frac{	heta }{2}}{1+	an \frac{	heta }{2}}$

$\Rightarrow$ $k=\frac{{(1+	an \frac{	heta }{2})}^{2}+{(1-	an \frac{	heta }{2})}^{2}}{(1-	an \frac{	heta }{2})(1+	an \frac{	heta }{2})}$

$\Rightarrow$ $k=\frac{1+{	an }^{2}\frac{	heta }{2}+2	an \frac{	heta }{2}+1+{	an }^{2}\frac{	heta }{2}-2	an \frac{	heta }{2}}{(1-	an \frac{	heta }{2})(1+	an \frac{	heta }{2})}$

$k=\frac{2(1+{	an }^{2}\frac{	heta }{2})}{(1-{	an }^{2}\frac{	heta }{2})}$

Using the identity $\cos 2x=\frac{1-{	an }^{2}x}{1+{	an }^{2}x}$ :

$k=\frac{2}{\cos 	heta }$

$\begin{gathered} 	heta ={\cos }^{-1}\frac{2}{3}\Rightarrow \cos 	heta =\frac{2}{3} \ k=\frac{2}{2/3}=3 \end{gathered}$

Solving the main equation :

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

Put $k=3$ :

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

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

Domain of ${\sin }^{-1}(3x-1)\Rightarrow 3x-1\in [-1,1]\Rightarrow x\in [0,\frac{2}{3}]$

Domain of ${\sin }^{-1}x\Rightarrow x\in [-1,1]$

Intersection domain of Equation 1 is $x\in [0,\frac{2}{3}]$

Take sine of both sides of Equation 1:

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

$\Rightarrow$ $3x-1=-\cos (2{\sin }^{-1}x)$

Using $\cos 2A=1-2{\sin }^{2}A$ :

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

$\Rightarrow$ $3x-1=2{x}^{2}-1$

$\Rightarrow$ $3x=2{x}^{2}\Rightarrow x(2x-3)=0$

$\Rightarrow$ $x=0x=\frac{3}{2}$

$\frac{3}{2}$ is outside the domain.

So, $x=0$ is the only solution.

Number of solutions $=1$.

Solutions · 1