The solution to the inequality $\operatorname{sen}^2 x - \frac{1}{2} \ge 0$ in the interval $[0, \pi]$ is:
$\operatorname{sen}^2 x \ge \frac{1}{2} \iff |\operatorname{sen} x| \ge \frac{\sqrt{2}}{2}$; 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:
$\operatorname{sen} x(1 + \operatorname{sen} x) > 0$. As $1 + \operatorname{sen} x \ge 0$, and it is null only when $\operatorname{sen} x = -1$, the inequality is equivalent to $\operatorname{sen} x > 0$.
Teste 174CESCEA-71
The solution of the inequality $\operatorname{sen}^2 x < 2\operatorname{sen} x$ in the closed interval $[0, 2\pi]$ is:
$\operatorname{sen} x(\operatorname{sen} x - 2) < 0$; as $\operatorname{sen} x - 2 < 0$ always, the inequality equals $\operatorname{sen} x > 0$: in $[0, 2\pi]$, $0 < x < \pi$.
Teste 175GV-72
The solution of the inequality $\sqrt{2}\cos^2 x > \cos x$ in the interval $[0, \pi]$ is:
$\cos x(\sqrt{2}\cos x - 1) > 0 \iff \cos x > \frac{\sqrt{2}}{2}$ or $\cos x < 0$; in $[0, \pi]$: $0 \le x < \frac{\pi}{4}$ or $\frac{\pi}{2} < x \le \pi$. Alternative (a) is correct, unless the extreme $\frac{\pi}{4}$, where equality holds (the inequality is strict).
Teste 176MACK-75
For $0 \le x \le 2\pi$, the solution set of $(\operatorname{sen} x + \cos x)^2 > 1$ is:
$(\operatorname{sen} x + \cos x)^2 = 1 + \operatorname{sen} 2x > 1 \iff \operatorname{sen} 2x > 0 \iff 0 < x < \frac{\pi}{2}$ or $\pi < x < \frac{3\pi}{2}$ (in $[0, 2\pi]$).
Teste 177ITA-76
The inequality $4\operatorname{sen}^2 x - 2(1 + \sqrt{2})\operatorname{sen} x + \sqrt{2} < 0$ has a solution $x$ such that:
The roots of the trinomial in $s = \operatorname{sen} x$ are $\frac{1}{2}$ and $\frac{\sqrt{2}}{2}$ (the discriminant is $4(\sqrt{2} - 1)^2$). The inequality applies to $\frac{1}{2} < \operatorname{sen} x < \frac{\sqrt{2}}{2}$, for example $30^\circ < x < 45^\circ$; every $x$ between $35^\circ$ and $45^\circ$ is a solution.
Teste 178MACK-73
If $0 \le \alpha \le \pi$ and, for all real $x$, $x^2 + x + \operatorname{tg}\alpha > \frac{3}{4}$, then:
The trinomial $x^2 + x + \left(\operatorname{tg}\alpha - \frac{3}{4}\right)$ is positive for all $x$ if the discriminant is negative: $1 - 4\operatorname{tg}\alpha + 3 < 0$, that is, $\operatorname{tg}\alpha > 1$. In $[0, \pi]$: $\frac{\pi}{4} < \alpha < \frac{\pi}{2}$.
Teste 179CESCEA-71
The solution of the inequality $\operatorname{sen} 2x \cdot \left(\sec^2 x - \frac{1}{3}\right) \le 0$ in the closed interval $[0, 2\pi]$ is:
Where is defined, $\sec^2 x - \frac{1}{3} \ge 1 - \frac{1}{3} > 0$; the inequality equates to $\operatorname{sen} 2x \le 0$ with $\cos x \ne 0$: $\frac{\pi}{2} < x \le \pi$ or $\frac{3\pi}{2} < x \le 2\pi$ (and also $x = 0$). The printed template indicates (b), which is the solution of $\operatorname{sen} 2x \ge 0$, that is, the opposite inequality.
Teste 180CESCEA-75
The values of $x \in\, ]0, \pi[$ for which $(1 + \operatorname{sen} x)(1 - \cos x)\left(\frac{\pi}{2} - x\right) < 0$ are such that:
In $]0, \pi[$, $1 + \operatorname{sen} x > 0$ and $1 - \cos x > 0$; the product is negative when $\frac{\pi}{2} - x < 0$, that is, $x > \frac{\pi}{2}$.
Teste 181MACK-73
The points of the trigonometric circle corresponding to the solutions of the system $\operatorname{sen} 2x > 0$, $\operatorname{cotg} x < 0$:
$\operatorname{sen} 2x = 2\operatorname{sen} x\cos x > 0$ requires sine and cosine of the same sign (1st or 3rd quadrant); $\operatorname{cotg} x < 0$ requires opposite signs (2nd or 4th). There is no solution.
Teste 182S. CARLOS-68
The inequality $|\cos x| \ge \operatorname{sen} x$, with $0 < x < 2\pi$, is valid if, and only if:
Where $\operatorname{sen} x \le 0$ ($\pi \le x < 2\pi$) the inequality automatically holds. Where $\operatorname{sen} x > 0$, $|\cos x| \ge \operatorname{sen} x$ is needed, that is, $0 < x \le \frac{\pi}{4}$ or $\frac{3\pi}{4} \le x < \pi$. Joining: $0 < x \le \frac{\pi}{4}$ or $\frac{3\pi}{4} \le x < 2\pi$.
Teste 183MAUÁ-69
All arcs between $0$ and $2\pi$ radians that satisfy the inequality $\cos x + \sqrt{3}\operatorname{sen} x > \sqrt{2}$ are included between:
$\cos x + \sqrt{3}\operatorname{sen} x = 2\operatorname{sen}\left(x + \frac{\pi}{6}\right) > \sqrt{2} \iff \frac{\pi}{4} < x + \frac{\pi}{6} < \frac{3\pi}{4} \iff \frac{\pi}{12} < x < \frac{7\pi}{12}$.
Teste 184ITA-71
Let $n$ be an integer, $n > 1$, and $x \in \left]0, \frac{\pi}{2}\right[$. Which statement below is always true?
It is Bernoulli’s inequality: for $0 < s < 1$, $(1 - s)^n \ge 1 - ns$ (proved by induction: $(1 - s)^{n+1} \ge (1 - ns)(1 - s) = 1 - (n + 1)s + ns^2 \ge 1 - (n + 1)s$).
Teste 185GV-75
For $y = \log(1 - \operatorname{sen}^2 x)$ to have real values, we must have, for $k$ integer:
$1 - \operatorname{sen}^2 x = \cos^2 x$ must be positive: $\cos x \ne 0$, that is, $x \ne \frac{\pi}{2} + k\pi$.
Teste 186MACK-74
Being $\operatorname{sen} x - \operatorname{sen} y = 2\operatorname{sen}\frac{x - y}{2}\cos\frac{x + y}{2}$ and remembering that $|\operatorname{sen} z| \le |z|$, $|\cos t| \le 1$ and $|ab| = |a||b|$, we can state that, for any real numbers $x$ and $y$:
It’s the same problem as the test 115: $|\operatorname{sen} x - \operatorname{sen} y| = 2\left|\operatorname{sen}\frac{x - y}{2}\right|\left|\cos\frac{x + y}{2}\right| \le |x - y|$.
Teste 187ITA-76
Regarding the product $P = (\operatorname{sen} bx + \operatorname{cossec} bx)(\cos bx + \sec bx)(\operatorname{tg} bx + \operatorname{cotg} bx)$ we can state that:
With $u = bx$: $P = \frac{\operatorname{sen}^2 u + 1}{\operatorname{sen} u} \cdot \frac{\cos^2 u + 1}{\cos u} \cdot \frac{1}{\operatorname{sen} u\cos u} = \frac{(\operatorname{sen}^2 u + 1)(\cos^2 u + 1)}{\operatorname{sen}^2 u\cos^2 u} > 0$, whenever it exists, that is, when $\operatorname{sen} u$ and $\cos u$ do not cancel each other out: $bx \ne \frac{k\pi}{2}$.
Teste 188ITA-68
Let $y = a^{\log\operatorname{tg} x}$, with $0 < a < 1$, where $\log u$ indicates the Neperian logarithm of $u$. Then $\log y \ge 0$ if:
$\log y = (\log\operatorname{tg} x)(\log a)$, with $\log a < 0$. Then $\log y \ge 0 \iff \log\operatorname{tg} x \le 0 \iff 0 < \operatorname{tg} x \le 1$: $0 < x \le \frac{\pi}{4}$ or $\pi < x \le \frac{5\pi}{4}$ (on the first lap).
