Question

please me with this!

Answer 1

$sin^{-1}(x) + cos^{-1}(-\frac{1}{2}) = \pi$
$sin[(sin^{-1}(x) + cos^{-1}(-\frac{1}{2})] = 0$
$sin(sin^{-1}(x)) \cdot cos(cos^{-1}(-\frac{1}{2})) + cos(sin^{-1}(x)) \cdot sin(cos^{-1}(-\frac{1}{2})) = 0$

$-\frac{x}{2} + \sqrt{1 - x^{2}} \cdot \sqrt{1 - (-\frac{1}{2})^{2}} = 0$
$\sqrt{1 - x^{2}} \cdot \sqrt{1 - (\frac{1}{4})} = \frac{x}{2}$
$\sqrt{1 - x^{2}} \cdot \sqrt{\frac{3}{4}} = \frac{x}{2}$

$\frac{\sqrt{3(1 - x^{2})}}{2} = \frac{x}{2}$
$\sqrt{3(1 - x^{2})} = x$
$3(1 - x^{2}) = x^{2}$
$3 - 3x^{2} = x^{2}$
$4x^{2} = 3$

$x^{2} = \frac{3}{4}$
$x = \pm \frac{\sqrt{3}}{2}$

Substitute both x-values in, and only $x = \frac{\sqrt{3}}{2}$ works.

Thus, $x = \frac{\sqrt{3}}{2}$ is your solution.