Assorted trigonometric, fractional and logarithmic inequalities

Text

Solve the following inequalities:

(a) $\left(\sin\dfrac x2-\dfrac{\sqrt2}{2}\right)\left(\cos x-\dfrac12\right)\ge0$

(b) $\dfrac{3\tan x-3}{\cos x-\dfrac12}\ge0$

(c) $\cos x-\sin x\le1$

(d) $\sin^2 x+3\cos^2 x+\sin x\cos x\le1$

(e) $\dfrac{2\log_3 x-2}{1-2^{x}}\le0$

Solution

(a) The product is $\ge0$ where the two factors have the same sign. Study the signs of $\sin\dfrac x2\ge\dfrac{\sqrt2}{2}$ (true for $\dfrac x2\in\left[\dfrac\pi4,\dfrac{3\pi}4\right]$, i.e. $x\in\left[\dfrac\pi2,\dfrac{3\pi}2\right]$) and of $\cos x\ge\dfrac12$ (true for $x\in\left[-\dfrac\pi3,\dfrac\pi3\right]$), then combine them with a sign chart.

(b) The fraction is $\ge0$ where numerator and denominator have the same sign. Numerator $3\tan x-3\ge0\iff\tan x\ge1$; denominator $\cos x-\dfrac12>0\iff\cos x>\dfrac12$. Build the sign chart accounting for the zeros and the existence conditions ($x\neq\dfrac\pi2+k\pi$ and $\cos x\neq\dfrac12$).

(c) With the linear formula $\cos x-\sin x=\sqrt2\cos\!\left(x+\dfrac\pi4\right)$, the inequality becomes $\sqrt2\cos\!\left(x+\dfrac\pi4\right)\le1$, i.e. $\cos\!\left(x+\dfrac\pi4\right)\le\dfrac{\sqrt2}{2}$, satisfied for $x+\dfrac\pi4\in\left[\dfrac\pi4,\dfrac{7\pi}4\right]$, i.e. $x\in\left[0,\dfrac{3\pi}2\right]$ (modulo $2\pi$).

(d) Using $\sin^2 x+\cos^2 x=1$ it rewrites as $1+2\cos^2 x+\sin x\cos x\le1$, i.e. $\cos x\left(2\cos x+\sin x\right)\le0$. Study the sign of the two factors $\cos x$ and $2\cos x+\sin x$ (the latter vanishes for $\tan x=-2$) with a sign chart.

(e) Domain $x>0$. For $x>0$ we have $2^x>1$, so the denominator $1-2^x<0$ is always negative; the fraction is $\le0$ when the numerator is $\ge0$: $2\log_3 x-2\ge0\iff\log_3 x\ge1\iff x\ge3$.

$\boxed{(c)\ \cos\!\left(x+\tfrac\pi4\right)\le\tfrac{\sqrt2}{2};\quad (d)\ \cos x\,(2\cos x+\sin x)\le0;\quad (e)\ x\ge3}$