the number of solution of 2^(cos x)=abs(sin x) in (0,pi) is

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- Jun 2nd 2011, 09:07 AMayushdadhwalnumber of solution
the number of solution of 2^(cos x)=abs(sin x) in (0,pi) is

- Jun 2nd 2011, 09:52 AMTheEmptySet
If you take the log of both sides of the equation you get

$\displaystyle \ln(2)\cos(x)=\ln(|sin(x)|)$

The sine function from zero to Pi over 2 goes from zero to 1 and is increasing on that interval. The graph is symmetric over the line x=PI/2.

This tells us that

$\displaystyle f(x)=\ln|\sin(x)|\le 0 $ increase from negative infinity to zero at x=Pi/2 and then decreases from 0 to negative infinite at Pi.

the function

$\displaystyle g(x)=\ln(2)\cos(x)$ has one zero at x =Pi/2 and decreases from ln(2) to negative ln(2) on the interval.

So this has one solution when x=Pi/2. Since both functions are monotone decreasing on Pi/2 Pi and

$\displaystyle \lim_{x \to \pi}g(x)=-\ln(2) \quad \lim_{x \to \pi}f(x)=-\infty$

They will cross exactly one more time in the inteval $\displaystyle \left( \frac{\pi}{2},\pi\right)$