- #1
Orion1
- 973
- 3
Show that the equation:
[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?
[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?