Question

Solve the following inequalities.
${\log }_{\frac{1}{2}}(2^{-x}-100\sin x)<x.$

Answer 1

${\log }_{\frac{1}{2}}(2^{-x}-100\sin x)<x$

$\Rightarrow \frac{\log (2^{-x}-100\sin x)}{\log \frac{1}{2}}<x$

$\Rightarrow \log (2^{-x}-100\sin x)<x(-\log 2)$

$\Rightarrow \log (2^{-x}-100\sin x)<\log 2^{-x}$

$\Rightarrow 2^{-x}-100\sin x<2^{-x}$

$\Rightarrow -100\sin x>0$

$\Rightarrow 100\sin x>0$

$\Rightarrow \sin x>0$

$\sin x$ is positive in first two quadrants. i.e., from $0$ to $\pi$

Also $\sin \left(x\right)$ has periodicity of $2\pi$, which mean it repeats after every $2\pi$

In general $\sin x>0$, if $xϵ(2n\pi ,(2n+1\left)\pi \right)$

where $n$ is any integer.