Solve

Question:

Solve $\cos ^{-1} \sqrt{3} x+\cos ^{-1} x=\frac{\pi}{2}$

Solution:

$\cos ^{-1} \sqrt{3} x+\cos ^{-1} x=\frac{\pi}{2}$

$\Rightarrow \cos ^{-1}\left[\sqrt{3} x \times x-\sqrt{1-(\sqrt{3} x)^{2}} \sqrt{1-x^{2}}\right]=\frac{\pi}{2}$                             $\left[\because \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)\right]$

$\Rightarrow \cos ^{-1}\left[\sqrt{3} x^{2}-\sqrt{1-3 x^{2}} \sqrt{1-x^{2}}\right]=\frac{\pi}{2}$

$\Rightarrow \sqrt{3} x^{2}-\sqrt{1-3 x^{2}} \sqrt{1-x^{2}}=\cos \frac{\pi}{2}$

$\Rightarrow \sqrt{3} x^{2}=\sqrt{1-3 x^{2}} \sqrt{1-x^{2}}$

$\Rightarrow 3 x^{4}=\left(1-3 x^{2}\right)\left(1-x^{2}\right)$

$\Rightarrow 3 x^{4}=1-3 x^{2}+3 x^{4}-x^{2}$

$\Rightarrow 4 x^{2}=1$

$\Rightarrow x^{2}=\frac{1}{4}$

$\Rightarrow x=\pm \frac{1}{2}$

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