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[Solved] Trigonometry roots

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Let   \large cos^{-1}\sqrt{x}=\Theta ,  then \large cos\Theta =\sqrt{x},      \large tan\Theta =\frac{sin\Theta }{cos\Theta }=\frac{\sqrt{1-cos^2\Theta }}{cos\Theta}=\frac{\sqrt{1-x}}{\sqrt{x}}.

Let      \large cot^{-1}\frac{1}{2}=\Gamma ,      then     \large cot\Gamma =\frac{1}{2},          \large sin\Gamma =\frac{1}{csc\Gamma }=\frac{1}{\sqrt{1+cot^2\Gamma }}=\frac{2}{\sqrt{5}} .

The equation is equal to

 \large \frac{\sqrt{1-x}}{\sqrt{x}}=\frac{2}{\sqrt{5}}\Rightarrow \frac{1-x}{x}=\frac{4}{5},

therefore \large x=\frac{5}{9} .

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