Trigonometric inequalities (chapter VIII). Teste 172CESCEA-74 The solution to the inequality $\operatorname{sen}^2 x - \frac{1}{2} \ge 0$ in the interval $[0, \pi]$ is: a) $0 \le x \le \frac{\pi}{4}$ or $\frac{3\pi}{4} \le x \le \pi$b) $\frac{\pi}{4} \le x \le \frac{3\pi}{4}$c) $\frac{\pi}{3} \le x \le \frac{2\pi}{3}$d) $0 \le x \le \frac{\pi}{6}$ or $\frac{5\pi}{6} \le x \le \pi$e) I don’t know $\operatorname{sen}^2 x \ge \frac{1}{2} \iff |\operatorname{sen} x| \ge \frac{\sqrt{2}}{2}$; in $[0, \pi]$, $\frac{\pi}{4} \le x \le \frac{3\pi}{4}$. Teste 173CESCEM-72 Consider the inequality $\operatorname{sen} x + \operatorname{sen}^2 x > 0$. It can be stated that: a) is only satisfied for $x$ in the first quadrantb) is only satisfied for $x$ between $0$ and $\pi$c) the inequality obtained by substituting $x$ for $-x$ is equivalent to the given inequalityd) the values of $x$ that satisfy it are precisely those for which $\operatorname{sen} x > 0$e) there is $x$ in the third quadrant that satisfies the inequality $\operatorname{sen} x(1 + \operatorname{sen} x) > 0$. As $1 + \operatorname{sen} x \ge 0$, and it is null only when $\operatorname{sen} x = -1$, the inequality is equivalent to $\operatorname{sen} x > 0$.
In each inequality, draw the cycle: mark the auxiliary line, highlight the appropriate arc and only then write the intervals, starting from $A$. Points that do not satisfy the strict inequality are open. Sine inequalities 271 Solve the inequality $0 \le \operatorname{sen} x < \dfrac{\sqrt{3}}{2}$. See the resolutionThe image of $x$ must be at the intersection of the cycle with the plane range between the straight lines $r$ ($v = 0$) and $s$ ($v = \frac{\sqrt{3}}{2}$), included $r$ and excluded $s$: Ative o JavaScript para ver esta figura interativa. $$2k\pi \le x < \frac{\pi}{3} + 2k\pi \quad\text{ou}\quad \frac{2\pi}{3} + 2k\pi < x \le \pi + 2k\pi.$$
For the tangent, the border is no longer a straight line parallel to an axis, but the straight line $OT$ that passes through the center: it cuts the cycle into two diametrically opposite points, and the solutions always come in pairs of arcs symmetric with respect to $O$ — the reflection of the tangent having a period $\pi$. The story The name tangent comes from the axis we use to measure it: the segment $\overline{AT}$ is on the tangent line to the cycle at $A$. The word, in Latin tangens, was introduced by the Danish mathematician Thomas Fincke in the book Geometria rotundi (1583), together with secans. Before, the Arabs already calculated this magnitude as the “shadow” of a vertical rod illuminated by the Sun, in a study linked to sundials (see chapter II).
For the cosine, the auxiliary line is vertical: the solutions for $\cos x > m$ are to the right of it, and those for $\cos x < m$, to the left. The story The interval notation we use in this book, with “turned” brackets to indicate an excluded extreme — $\left]a, b\right[$ — is French tradition; it was disseminated in the 20th century by the group Nicolas Bourbaki, a collective pseudonym of French mathematicians who rewrote the foundations of mathematics in a series of books. In English-speaking countries, notation with parentheses, $(a, b)$, is more common.
Let $f$ and $g$ be two trigonometric functions of the variable $x$. Solving the inequality $f(x) < g(x)$ is to obtain the set $S$ — the solution set or truth set — of numbers $r$ for which $f(r) < g(r)$ is a true sentence. The story The signals $<$ and $>$ appeared printed in 1631, in the book Artis analyticae praxis, published ten years after the death of the English mathematician Thomas Harriot, who was also an astronomer and navigator. The signs $\le$ and $\ge$ (with a dash under the sign) were introduced in 1734 by the French Pierre Bouguer, mathematician and hydrographer. Before that, inequalities were written out in full — “is greater than”, “is less than”.