If ^-1(23x)+^-1(34x)=2, x34, then x is equal to:

Question

If $\cos^{-1}\left(\dfrac{2}{3x}\right)+\cos^{-1}\left(\dfrac{3}{4x}\right)=\dfrac{\pi}{2}$, $x>\dfrac34$, then $x$ is equal to:

Accepted Answer

$$\cos^{-1}\frac{2}{3x}=\frac{\pi}{2}-\cos^{-1}\frac{3}{4x}=\sin^{-1}\frac{3}{4x}.$$

Let $\sin^{-1}\dfrac{3}{4x}=\theta$, so $\sin\theta=\dfrac{3}{4x}$, $\cos\theta=\sqrt{1-\dfrac{9}{16x^2}}=\dfrac{\sqrt{16x^2-9}}{4x}$, so $\theta=\cos^{-1}\left(\dfrac{\sqrt{16x^2-9}}{4x}\right)$.

$$\cos^{-1}\frac{2}{3x}=\cos^{-1}\left(\frac{\sqrt{16x^2-9}}{4x}\right)\ \Rightarrow\ \frac{2}{3x}=\frac{\sqrt{16x^2-9}}{4x}\ \Rightarrow\ \frac{8}{3}=\sqrt{16x^2-9}\ \Rightarrow\ \frac{64}{9}+9=16x^2.$$

$$x^2=\frac{64+81}{9\times16}=\frac{145}{144}\ \Rightarrow\ x=\frac{\sqrt{145}}{12}\quad(x>0).$$

Solutions · 1