Trigonometric equations and solutions on an interval (chapter VII).
Teste 122FUVEST-77
In the range $\frac{\pi}{2} \le x \le \pi$, the equation $\sqrt{1 - \operatorname{sen}^2 x} + \cos x = -\sqrt{2}$:
$\sqrt{1 - \operatorname{sen}^2 x} = |\cos x|$, and in the 2nd quadrant $|\cos x| = -\cos x$. The first member is worth $-\cos x + \cos x = 0 \ne -\sqrt{2}$.
Teste 123PUC-76
The values of $x$ that satisfy the equation $\cos\left(3x - \frac{\pi}{5}\right) = 0$ are ($k = 0, \pm 1, \pm 2, \ldots$):
The smallest positive value of $x$ for which $9^{-\cos x} = \frac{1}{3}$ is:
$9^{-\cos x} = 3^{-2\cos x} = 3^{-1} \Rightarrow \cos x = \frac{1}{2}$, whose smallest positive value is $\frac{\pi}{3}$.
Teste 125CESCEM-73
If the point $(x_0; y_0)$ belongs to the graph of the function $y = \operatorname{tg} x$, then a necessary and sufficient condition for the point $(a; y_0)$ to also belong to this graph is:
$\operatorname{tg} a = \operatorname{tg} x_0 \iff a = x_0 + k\pi$: the difference $a - x_0$ must be a multiple of $\pi$ (the tangent period).
Teste 126EESCUSP-68
The solutions to the equation $\operatorname{sen}\pi x = \operatorname{sen}[\pi(2x + 1)]$ are of the form:
$\operatorname{sen}\alpha = \operatorname{sen}\beta$ gives: $\pi(2x + 1) = \pi x + 2k\pi$, that is, $x = 2k - 1$ (odd integers); or $\pi(2x + 1) = \pi - \pi x + 2k\pi$, that is, $x = \frac{2k}{3}$. No alternative exactly describes this set: (a), for example, includes $x = \frac{1}{3}$, which is not a solution.
Teste 127ITA-77
Solving the equation $\operatorname{tg}\left(2\log x - \frac{\pi}{6}\right) - \operatorname{tg}\left(\log x + \frac{\pi}{3}\right) = 0$ (natural logarithm), we have ($k = 0, 1, 2, \ldots$):
With $L = \log x$ (natural): $\operatorname{tg}\left(2L - \frac{\pi}{6}\right) = \operatorname{tg}\left(L + \frac{\pi}{3}\right) \Rightarrow 2L - \frac{\pi}{6} = L + \frac{\pi}{3} + k\pi \Rightarrow L = \frac{\pi}{2} + k\pi$ (and the tangents exist, as the arcs are worth $\frac{5\pi}{6} + k\pi$). Therefore $x = e^{\pi/2 + k\pi}$, $k \in \mathbb{Z}$, which is the set of (b).
Teste 128CESGRANRIO-77
The number of roots of the equation $\cos x + \operatorname{sen} x = 0$ in the interval $[\pi, 3\pi]$ is:
$\cos x + \operatorname{sen} x = 0 \iff \operatorname{tg} x = -1 \iff x = \frac{3\pi}{4} + k\pi$. In $[\pi, 3\pi]$: $\frac{7\pi}{4}$ and $\frac{11\pi}{4}$.
Teste 129MACK-75
If $\operatorname{tg} 4x + \operatorname{tg}\left(2x - \frac{\pi}{4}\right) = 0$ to $0 < x < \frac{\pi}{2}$, then $x$ can be equal to:
$\operatorname{tg} 4x = -\operatorname{tg}\left(2x - \frac{\pi}{4}\right) = \operatorname{tg}\left(\frac{\pi}{4} - 2x\right) \Rightarrow 4x = \frac{\pi}{4} - 2x + k\pi \Rightarrow x = \frac{\pi}{24} + \frac{k\pi}{6}$. In $\left]0, \frac{\pi}{2}\right[$: $\frac{\pi}{24}$, $\frac{5\pi}{24}$ and $\frac{9\pi}{24}$. Of the alternatives, only $\frac{5\pi}{24}$ works.
Teste 130CESCEA-70
If $a$ is the smallest positive root of the equation $(\operatorname{tg} x - 1)(4\operatorname{sen}^2 x - 3) = 0$, then the value of $\operatorname{sen}^4 a - \cos^2 a$ is:
Positive roots come from $\operatorname{tg} x = 1$ ($x = \frac{\pi}{4}, \ldots$) or $\operatorname{sen} x = \pm\frac{\sqrt{3}}{2}$ ($x = \frac{\pi}{3}, \ldots$). The smallest is $a = \frac{\pi}{4}$: $\operatorname{sen}^4 a - \cos^2 a = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.
Teste 131CESCEA-75
The sum of the roots of the equation $1 - 4\cos^2 x = 0$ between $0$ and $\pi$ is:
$\cos^2 x = \frac{1}{4} \Rightarrow \cos x = \pm\frac{1}{2}$; between 0 and $\pi$: $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, sum $\pi$.
Teste 132CESGRANRIO-76
In the interval $[0, 6\pi]$, the trigonometric equation $\cos 2x + 2\operatorname{sen}^2 x + 2 = 0$:
$\cos 2x + 2\operatorname{sen}^2 x = 1$ for all $x$; the first member is always worth 3.
Teste 133ITA-69
The equation $\operatorname{sen}^2\frac{3x}{2} - \cos\frac{3x}{2} = a$ has a solution for particular values of $a$. Select the correct item:
With $c = \cos\frac{3x}{2} \in [-1, 1]$: $a = 1 - c^2 - c$, which ranges from $-1$ (in $c = 1$) to $\frac{5}{4}$ (in $c = -\frac{1}{2}$). There is a solution for $-1 \le a \le \frac{5}{4}$; the only offered range contained in this is that of (c).
Teste 134CESCEM-73
The values of $x$ between $0$ and $2\pi$ that satisfy the equation $2\operatorname{sen}^2 x + |\operatorname{sen} x| - 1 = 0$ are:
With $u = |\operatorname{sen} x| \ge 0$: $2u^2 + u - 1 = 0 \Rightarrow u = \frac{1}{2}$ (the root $-1$ is discarded). $|\operatorname{sen} x| = \frac{1}{2}$ gives the four values of (b).
Teste 135GV-75
The solution of the equation $\dfrac{625^{\cos^2 x}}{25^{\cos x}} = 1$ to $0 \le x < \frac{\pi}{2}$ is:
$\frac{625^{\cos^2 x}}{25^{\cos x}} = 25^{2\cos^2 x - \cos x} = 1 \Rightarrow \cos x(2\cos x - 1) = 0$. In $\left[0, \frac{\pi}{2}\right[$, $\cos x \ne 0$; remains $\cos x = \frac{1}{2}$, $x = \frac{\pi}{3}$.
Teste 136GV-73
Given the equation $\cos^2 x - 2\sec^2 x = 1$, with $0 \le x \le \pi$, then:
With $c = \cos^2 x$: $c - \frac{2}{c} = 1 \Rightarrow c^2 - c - 2 = 0 \Rightarrow c = 2$ or $c = -1$, both impossible for a squared cosine.
Teste 137GV-75
The set of all solutions to the equation $\cos^2 x \cdot \operatorname{tg} x = \operatorname{sen} x$ is the set of numbers $x$ such that $x$ is equal to:
With $\cos x \ne 0$: $\cos^2 x \cdot \frac{\operatorname{sen} x}{\cos x} = \operatorname{sen} x \Rightarrow \operatorname{sen} x(\cos x - 1) = 0$. The two possibilities ($\operatorname{sen} x = 0$ or $\cos x = 1$) give $x = k\pi$.
Teste 138CESCEM-73
As a function of a relative integer $k$, all solutions of the equation $\cos^4\alpha + \operatorname{sen}^4\alpha - 2\operatorname{sen}^2\alpha\cos^2\alpha = 1$ are given by:
The first member is $(\cos^2\alpha - \operatorname{sen}^2\alpha)^2 = \cos^2 2\alpha$. Then $\cos^2 2\alpha = 1 \Rightarrow 2\alpha = k\pi \Rightarrow \alpha = \frac{k\pi}{2}$. The printed template indicates (b), $\alpha = k\pi$; but $\alpha = \frac{\pi}{2}$ is also a solution ($0 + 1 - 0 = 1$), and (b) only gives part of the solutions.
Teste 139CESCEM-72
The expression $\operatorname{sen}^6 x + \cos^6 x = 1 - 3\operatorname{sen}^2 x\cos^2 x$:
$\operatorname{sen}^6 x + \cos^6 x = (\operatorname{sen}^2 x + \cos^2 x)^3 - 3\operatorname{sen}^2 x\cos^2 x(\operatorname{sen}^2 x + \cos^2 x) = 1 - 3\operatorname{sen}^2 x\cos^2 x$ to all $x$ (see chapter VII).
Teste 140ITA-71
What is the smallest value of $x$ that checks the equation $\operatorname{tg} x + 3\operatorname{cotg} x = 3$?
With $t = \operatorname{tg} x$: $t + \frac{3}{t} = 3 \Rightarrow t^2 - 3t + 3 = 0$, from discriminant $9 - 12 < 0$. There is no solution.
Teste 141MACK-77
The two acute angles of a non-isosceles right triangle are roots of the equation (in $x$) $3\operatorname{tg} x + k^2\operatorname{cotg} x = 4k$. Then:
Multiplying by $\operatorname{tg} x$: $3t^2 - 4kt + k^2 = 0$, that is, $(3t - k)(t - k) = 0$, with roots $k$ and $\frac{k}{3}$. Acute angles are complementary, so the product of tangents is 1: $\frac{k^2}{3} = 1$, $k = \sqrt{3}$ (tangents are positive).
Teste 142UNB-74
If $\sec^2 x + \operatorname{tg} x - 7 = 0$ and $0 < x < \frac{\pi}{2}$, then:
$1 + \operatorname{tg}^2 x + \operatorname{tg} x - 7 = 0 \Rightarrow \operatorname{tg} x = 2$ (in the 1st quadrant). Then $\sec^2 x = 5$ and $\cos x = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$.
Teste 143CESCEA-71
The solution set of the equation $3\operatorname{tg}^2 x + 5 = \dfrac{7}{\cos x}$ in the closed interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is:
With $s = \sec x$: $3(s^2 - 1) + 5 = 7s \Rightarrow 3s^2 - 7s + 2 = 0 \Rightarrow s = 2$ or $s = \frac{1}{3}$ (impossible). $\cos x = \frac{1}{2}$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ gives $\pm\frac{\pi}{3}$.
Teste 144FEI-73
The equation $\operatorname{sen} 2x = \operatorname{sen} x$, in the interval $-\frac{\pi}{4} \le x \le \frac{5\pi}{4}$, has:
$\operatorname{sen} 2x = \operatorname{sen} x \iff \operatorname{sen} x(2\cos x - 1) = 0$. In the range: $x = 0$, $x = \pi$ and $x = \frac{\pi}{3}$ ($-\frac{\pi}{3}$ and $\frac{5\pi}{3}$ are left out). The printed template indicates 4 solutions, but $-\frac{\pi}{3} < -\frac{\pi}{4}$ and $\frac{5\pi}{3} > \frac{5\pi}{4}$ are out of range.
Teste 145ITA-72
Choose a solution to the trigonometric equation $\sqrt{3}\operatorname{sen} x + \cos x = \sqrt{3}$:
$\sqrt{3}\operatorname{sen} x + \cos x = 2\operatorname{sen}\left(x + \frac{\pi}{6}\right) = \sqrt{3}$ gives $x + \frac{\pi}{6} = \frac{\pi}{3} + 2k\pi$ or $\frac{2\pi}{3} + 2k\pi$, that is, $x = \frac{\pi}{6} + 2k\pi$ or $x = \frac{\pi}{2} + 2k\pi$. The template indicates (b); Note, however, that (d) also gives solutions to the equation — the question has two correct alternatives.
Teste 146ITA-73
Let the equation be $(\log_e m)\operatorname{sen} x \pm \cos x = \log_e m$. What are the conditions on $m$ for the equation to admit a solution?
It takes $m > 0$ to exist $\log_e m$. For all $m > 0$ the equation has a solution: $x = \frac{\pi}{2} + 2k\pi$ always works ($L = L$). The alternatives impose constraints that do not exist — in (a), for example, $m = 1$ is excluded, but with $m = 1$ there is still the solution $x = \frac{3\pi}{2}$.
Teste 147PUC-73
If $\dfrac{1 - \operatorname{tg} x}{1 + \operatorname{tg} x} = 1 - \operatorname{sen} 2x$, then the values of $x$ are ($k = 0, \pm 1, \pm 2, \ldots$):
The first member is $\frac{\cos x - \operatorname{sen} x}{\cos x + \operatorname{sen} x}$ and the second is $(\cos x - \operatorname{sen} x)^2$. Then $\cos x = \operatorname{sen} x$ ($x = \frac{\pi}{4} + k\pi$, both members null) or $(\cos x - \operatorname{sen} x)(\cos x + \operatorname{sen} x) = \cos 2x = 1$ ($x = k\pi$). The template indicates (a), which gives only one of the two families: no alternative provides the complete solution set.
In the interval $0 \le x \le 2\pi$, the number of solutions of the trigonometric equation $\cos^9 x + \cos^8 x + \cos^7 x + \cdots + \cos x + 1 = 0$:
With $c = \cos x$, the first member is the sum of a geometric progression: $\frac{c^{10} - 1}{c - 1}$ (if $c \ne 1$; if $c = 1$ it is worth 10). It is zero if $c^{10} = 1$ and $c \ne 1$, that is, $c = -1$: only $x = \pi$.
Teste 150ITA-71
The equation $\operatorname{sen}(\cos x) \cdot \cos(\cos x) = 1$ is satisfied for:
Being $0 < x < \pi$, the equation $2\log\operatorname{sen} x + \log 2 = 0$ has the solution:
$\log(2\operatorname{sen}^2 x) = 0 \Rightarrow \operatorname{sen}^2 x = \frac{1}{2}$; in $]0, \pi[$, $\operatorname{sen} x = \frac{\sqrt{2}}{2}$ ($\log\operatorname{sen} x$ requires positive sine): $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.
Teste 152ITA-72
What conditions must $a$ and $k$ satisfy for the equality $\log(\sec a) = k$ to make sense?
The logarithm requires $\sec a > 0$, that is, $\cos a > 0$: $-\frac{\pi}{2} + 2n\pi < a < \frac{\pi}{2} + 2n\pi$; in this case $\sec a \ge 1$ and $k = \log\sec a \ge 0$. Alternative (a) gives only one of these ranges ($n = 0$); as a condition for equality to make sense, it is sufficient but not necessary — that is why the template indicates (e).
Teste 153MACK-77
The number of real solutions of the equation $x^2 - x - \cos x = 0$, with $-\pi \le x \le \pi$, is:
Let $f(x) = x^2 - x - \cos x$. We have $f(-\pi) > 0$, $f(0) = -1 < 0$ and $f(\pi) > 0$: there is at least one root on each side of 0. Like $f''(x) = 2 + \cos x > 0$, $f$ is convex and has no more than two roots. (Graphically: the parabola $y = x^2 - x$ cuts the curve $y = \cos x$ at two points.)
Teste 154SANTA CASA-77
The smallest value of $x$ that satisfies the equation $\log x = \cos x$ is between:
For $x \le 1$, $\log x \le 0 < \cos x$ (if $x$ is between 0 and 1). In $x = 1$, $\log 1 = 0 < \cos 1 \approx 0{,}54$; in $x = 1{,}6$, $\log 1{,}6 > 0 > \cos 1{,}6$. The first intersection is between 1 and 1,6 (about 1,42 in base 10, or 1,30 in base $e$).
Teste 155ITA-71
Given the equation $\log(\cos x) = \operatorname{tg} x$, the solutions of this equation in $x$ satisfy the relation:
It takes $\cos x > 0$, and then $\log(\cos x) \le 0$; logo $\operatorname{tg} x \le 0$. The solutions are in the 4th quadrant (including multiples of $2\pi$, such as $x = 2\pi$, where both members are null). In the first positive turn, this is $\frac{3\pi}{2} < x \le 2\pi$.
Teste 156ITA-76
Solving the equation $3\operatorname{sen}^2(e^x) - 2\sqrt{3}\operatorname{sen}(e^x)\cos(e^x) - 3\cos^2(e^x) = 0$, we obtain:
Dividing by $\cos^2(e^x)$: $3t^2 - 2\sqrt{3}t - 3 = 0$, with $t = \operatorname{tg} e^x$, from roots $\sqrt{3}$ and $-\frac{\sqrt{3}}{3}$. Then $e^x = \frac{\pi}{3} + k\pi$ or $e^x = -\frac{\pi}{6} + k\pi$, which together form $e^x = \frac{k\pi}{2} - \frac{\pi}{6}$, $k = 1, 2, 3, \ldots$ (you need $e^x > 0$).
Teste 157ITA-73
Let the equation be $3\operatorname{tg} 3x = [3(\log_e t)^2 - 4\log_e t + 2]\operatorname{tg} x$, $x \ne n\pi$. What are the conditions on $t$ for the equation to admit a solution?
With $u = \operatorname{tg} x \ne 0$ and $A = 3L^2 - 4L + 2$ ($L = \log_e t$): $\frac{3(3u - u^3)}{1 - 3u^2} = Au \Rightarrow u^2 = \frac{A - 9}{3(A - 1)}$, which needs to be positive: $A < 1$ or $A > 9$. $A < 1 \iff \frac{1}{3} < L < 1 \iff e^{1/3} < t < e$; $A > 9 \iff L < -1$ or $L > \frac{7}{3} \iff 0 < t < \frac{1}{e}$ or $t > e^{7/3}$.
