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[tex]2x - 1 - \sin x = 0[/tex]

has exactly one real root.

[tex]\frac{d}{dx} (2x - 1 - \sin x) = 2 - \cos x = 0[/tex]

[tex]2 = \cos x[/tex]

[tex]x = \cos^{-1} 2[/tex]

Is there a better way to approach the root?

any suggestions?

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# Real Root Help!

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