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How to find $x$ for $1 + \sin(x/2) = \cos x$ ?

From the equation, I can figure out that it is satisfied at $x = 0$ by looking. How do I find the other solutions to this equation?

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Do you know about double-angle formulas? – cardinal Apr 17 '11 at 4:05

Re-writing this as $\cos(2y) = 1 + \sin(y)$ for $y = \frac{x}{2}$ and using the fact that $\sin^2(y) = \frac{1 - \cos(2y)}{2}$, we can simplify the equation to $1 - 2\sin^2(y) = 1 + \sin(y)$ and thus $2\sin^2(y) + \sin(y) = (2\sin(y) + 1)\sin(y) = 0$. This means either $\sin(y) = 0$ or $\sin(y) = \frac{-1}{2}$, which has solutions of the form $y = n\pi$ or $y = \frac{12n\pi - \pi}{6}$ or $\frac{12n\pi + 7\pi}{6}$ for $n$ an integer. Clearly $x = 2y$.

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You might edit your post to make convert $sin$ to $\sin$ and $cos$ to $\cos$ by placing a backslash before each one. – cardinal Apr 17 '11 at 4:20
Thanks for the tip, I had thought you needed to use \text{}. – Alex Becker Apr 17 '11 at 4:39
Sure. That works for $\log$, $\exp$, $\min$, $\max$, etc., i.e., nearly all the "standard" functions. – cardinal Apr 17 '11 at 4:43
I think you meant $y=x/2$, not $y=2x$. – Hans Lundmark Apr 17 '11 at 10:20
@Hans: Oops, yes. – Alex Becker Apr 17 '11 at 22:53

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