Answer
$x=1$
Work Step by Step
Step 1. Let $u=sin^{-1}x, v=cos^{-1}x$, we have $sin(u)=x, cos(v)=x$, take the cosine of each side of the equation, we get $cos(2u+v)=cos(\pi)$.
Step 2. Use the Addition Formula for the above, we get $cos(2u)cos(v)-sin(2u)sin(v)=-1$
Step 3. Use the Double-Angle Formula, we get $(1-2sin^2u)cos(v)-2sin(u)cos(u)sin(v)=-1$
Step 4. Based on Step 1, $sin(u)=x, cos(u)=\pm\sqrt {1-x^2}, cos(v)=x, sin(v)=\pm\sqrt {1-x^2}$
Step 5. Plug-in the above results into the equation from Step 3, we get
$(1-2x^2)x\pm2x\sqrt {1-x^2}\sqrt {1-x^2}=-1$ which leads to $x-2x^3\pm(2x-2x^3)=-1$ or
$4x^3-3x-1=0$ and $-x=-1$
Step 6. Factor the first equation as $3x^3-3x+x^3-1=3x(x^2-1)+(x-1)(x^2+x+1)=(x-1)(3x^2+3x+x^2+x+1)=(x-1)(4x^2+4x+1)=(x-1)(2x+1)^2=0$, thus we have the solutions as $x=-0.5, 1$, check these numbers with the original equation, only one is the real solution $x=1$
Extra: the figure shows the graphical solutions of the equation.
