Trigonometric inequalities with sign analysis

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Solve the following inequalities:

(a) $\dfrac12-\sin x\le0$

(b) $3\tan x-\sqrt3\le0$

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

(d) $\dfrac{\dfrac{\sqrt3}{2}-\sin x}{\dfrac{\sqrt3}{2}+\cos x}\le0$

(e) $\sin x\left[\cos\dfrac x2-\dfrac12\right]\ge0$

(f) $\cos x-\sqrt3\sin x\le1$

(g) $\sqrt2\cos x+\sin x\ge\sqrt3$

(h) $2\sin^2 x-3\cos^2 x-4\sin x\cos x\ge0$

Solution

(a) This is equivalent to $\sin x\ge\dfrac12$, so $x\in\left[\dfrac\pi6+2k\pi,\ \dfrac{5\pi}6+2k\pi\right]$.

(b) This is equivalent to $\tan x\le\dfrac{\sqrt3}{3}$: on each period $\left(-\dfrac\pi2,\dfrac\pi2\right)$ it holds for $x\in\left(-\dfrac\pi2,\ \dfrac\pi6\right]$, repeated with period $\pi$.

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

(d) The fraction is $\le0$ where numerator and denominator have opposite signs (or the numerator vanishes): $\sin x\le\dfrac{\sqrt3}2$ makes the numerator $\ge0$, while $\cos x>-\dfrac{\sqrt3}2$ makes the denominator $>0$; conclude with a sign study over the full turn.

(e) $\cos\dfrac x2\ge\dfrac12 \iff x\in\left[-\dfrac{2\pi}3,\dfrac{2\pi}3\right]$; this interval must be combined with the sign of $\sin x$ using a sign chart.

(f) Using the linear-combination formula for sine and cosine, $\cos x-\sqrt3\sin x=2\cos\!\left(x+\dfrac\pi3\right)$, so the inequality becomes $2\cos\!\left(x+\dfrac\pi3\right)\le1$, i.e. $\cos\!\left(x+\dfrac\pi3\right)\le\dfrac12$.

(g) The amplitude is $\sqrt{(\sqrt2)^2+1^2}=\sqrt3$, so $\sqrt2\cos x+\sin x=\sqrt3\cos(x-\varphi)$ with $\tan\varphi=\dfrac1{\sqrt2}$. The inequality $\sqrt3\cos(x-\varphi)\ge\sqrt3$ is equivalent to $\cos(x-\varphi)\ge1$, which holds only at the absolute maximum: $x=\varphi+2k\pi$.

(h) This is a homogeneous second-degree inequality in sine and cosine. Applying the double-angle formulas it rewrites as $-\dfrac12-\dfrac52\cos2x-2\sin2x\ge0$, which reduces to a linear inequality in $\sin2x$ and $\cos2x$ solvable with the auxiliary-angle method.

$\boxed{(a)\ x\in\left[\frac\pi6,\frac{5\pi}6\right];\ \ (b)\ \tan x\le\frac{\sqrt3}3;\ \ (f)\ \cos\!\left(x+\frac\pi3\right)\le\frac12;\ \ (g)\ x=\varphi+2k\pi,\ \tan\varphi=\frac1{\sqrt2};\ \ (h)\ -\frac12-\frac52\cos2x-2\sin2x\ge0}$