system of equations involving inverse function

Question

The number of ordered pairs $(x,y)$ which satisfy the system of equations

$(\cos^{-1} x)^2+\sin^{-1}(y)=1$ and

$\cos^{-1}(x)+(\sin^{-1}y)^2=1$ are

Try: Adding these two equations.

we have

$\displaystyle (\cos^{-1} x)^2+\cos^{-1}x+(\sin^{-1} y)^2+\sin^{-1}y=2$

$\displaystyle \bigg(\cos^{-1} x+\frac{1}{2}\bigg)^2+\bigg(\sin^{-1}y+\frac{1}{2}\bigg)^2=\frac{3}{2}.\cdots (1)$

And subtracting equation

we have $\displaystyle (\cos^{-1} x)^2-\cos^{-1}x+(\sin^{-1} y)^2-\sin^{-1}y=0$

$\displaystyle \bigg(\cos^{-1} x-\frac{1}{2}\bigg)^2+\bigg(\sin^{-1}y-\frac{1}{2}\bigg)^2=0\cdots (2)$

Now i dont got any clue how to ho ahead from that points.

could some help me thanks

Answer 1

$\cos^{-1}x=a,\sin^{-1}y=b$

We have $a^2+b=1=a+b^2$

$\implies0=a^2-b^2-(a-b)=(a-b)(a+b-1)$

If $a+b-1=0\iff b=1-a, 1=a^2+b=a^2+1-a\implies a(a-1)=0$

If $a=b, 1=a^2+b=a^2+a\implies b=a=\dfrac{-1\pm\sqrt5}2$

$\cos^{-1}x=a=\dfrac{-1\pm\sqrt5}2, x=?$

Answer 2

Hints: if you put $a=\cos ^{-1}x$ and $b=\cos ^{-1}y$ the you get $a+b^{2}=1$ and $b+a^{2}=1$. Plug in the value of $a$ from the first equation into the second and you get $b(b^{3}-2b+1)=0$. This can be written as $b(b-1)(b^{2}+b-1)=0$. Solve the quadratic equation you can find all possible solutions.