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Inverse trig function question (1 Viewer)

Thread starter 720YoloSwagScope
Start date Jul 25, 2015

720YoloSwagScope

Member

Jul 25, 2015

How do I solve this?

Help is appreciated thanks!

$$Let $\alpha=\cos^{-1}{x} \to x=\cos{\alpha}$. Drawing a right-angled triangle, you find that $\sin{\alpha}=\sqrt{1-x^2}\to\alpha=\sin^{-1}{\sqrt{1-x^2}}$. Therefore, $\cos^{-1}{x}=\sin^{-1}{\sqrt{1-x^2}}$. Thus, the equation can be written as: \\$

$\begin{align*}\sin^{-1}{2x-1}&=\sin^{-1}{\sqrt{1-x^2}}\\2x-1&=\sqrt{1-x^2} \\\textup{Squaring both sides: }4x^2-4x+1&=1-x^2\\5x^2-4x&=0 \\x(5x-4)&=0 \\\therefore x&=0, \frac{4}{5}\end{align*}$

$$If you test solutions you find that $x=\frac{4}{5}$ is the only solution.$