Question

Solve for x : $si{n}^{-1}\frac{3x}{5}+si{n}^{-1}\frac{4x}{5}=si{n}^{-1}x$
$x=0,1,-1$

Answer 1

$si{n}^{-1}\frac{3x}{5}+si{n}^{-1}\frac{4x}{5}=si{n}^{-1}xsi{n}^{-1}\frac{3x}{5}=si{n}^{-1}x-si{n}^{-1}\frac{4x}{5}si{n}^{-1}\frac{3x}{5}=si{n}^{-1}x[x\sqrt{\frac{1-16x^2}{25}}-\frac{4x}{5}\sqrt{1-x^2}]...\left(1\right)\frac{3x}{5}=x\sqrt{\frac{1-16x^2}{25}}-\frac{4x}{5}\sqrt{1-x^2}=\frac{x}{5}\sqrt{25-16x^2}-\frac{4x}{5}\sqrt{1-x^2}3=\sqrt{25-16x^2}-4\sqrt{1-x^2}9=(25-16x^2)-8\sqrt{(25-16x^2)(1-x^2)+16(1-x^2)}9=41-32x^2-8\sqrt{(25-16x^2)(1-x^2)}32x^2-32=-8\sqrt{(25-16x^2)(1-x^2)}4-4x^2=\sqrt{(25-16x^2)(1-x^2)}16(1-x^2{)}^2=(25-16x^2\left)\right(1-x^2\left)16\right(1-x^2)=25-16x^{2}or1-x^2=016-16x^2=25-16x^2{x}^2=1x=0x=\pm 1$
$16(1-x^2{)}^2=(25-16x^2\left)16\right(1-x^2)=25-16x^{2}16-16x^2=25-16x^{2}x=0$