Exercises from chapter VIII — Inequalities

The 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 $s$:

$|\operatorname{sen} x| \ge \frac{\sqrt{3}}{2} \iff \operatorname{sen} x \le -\frac{\sqrt{3}}{2}$ or $\operatorname{sen} x \ge \frac{\sqrt{3}}{2}$. The image of $x$ must be at the intersection of the cycle with the half-plane below $r$ ($v = -\frac{\sqrt{3}}{2}$) or with the half-plane above $s$ ($v = \frac{\sqrt{3}}{2}$):

which is summarized in $\frac{\pi}{3} + k\pi \le x \le \frac{2\pi}{3} + k\pi$.

$2\operatorname{sen}^2 x < \operatorname{sen} x \iff 2\operatorname{sen}^2 x - \operatorname{sen} x < 0 \iff \operatorname{sen} x\,(2\operatorname{sen} x - 1) < 0 \iff 0 < \operatorname{sen} x < \frac{1}{2}$.

Examining the trigonometric cycle:

Making $2x = y$, we have the inequality $\operatorname{sen} y > 0$. Examining the cycle, $2k\pi < y < \pi + 2k\pi$. As $x = \frac{y}{2}$, it results

But $x \in [0, 2\pi]$; then only solutions with $k = 0$ or $k = 1$ are of interest:

• $k = 0$: $0 < x < \frac{\pi}{2}$;
• $k = 1$: $\pi < x < \frac{3\pi}{2}$.

The image of $x$ must be at the intersection of the cycle with the plane range between the straight lines $s$ ($u = -\frac{3}{2}$) and $r$ ($u = 0$). As the line $s$ does not even touch the cycle, the condition $\cos x \ge -\frac{3}{2}$ is always valid, and $\cos x \le 0$ remains:

Examining the trigonometric cycle:

Transforming the sum into a product:

Doing $x - \frac{\pi}{4} = y$, we have $\cos y \ge \frac{1}{2}$; examining the cycle, $2k\pi \le y \le \frac{\pi}{3} + 2k\pi$ or $\frac{5\pi}{3} + 2k\pi \le y < 2\pi + 2k\pi$. Like $x = y + \frac{\pi}{4}$:

that is, in a more compact form, $-\frac{\pi}{12} + 2k\pi \le x \le \frac{7\pi}{12} + 2k\pi$.

Doing $3x = y$, we have $\cos y \le \frac{1}{2}$; examining the cycle, $\frac{\pi}{3} + 2k\pi \le y \le \frac{5\pi}{3} + 2k\pi$. Like $x = \frac{y}{3}$:

For $x \in [0, 2\pi]$ only $k = 0$, $1$ or $2$ are of interest:

• $k = 0$: $\frac{\pi}{9} \le x \le \frac{5\pi}{9}$;
• $k = 1$: $\frac{7\pi}{9} \le x \le \frac{11\pi}{9}$;
• $k = 2$: $\frac{13\pi}{9} \le x \le \frac{17\pi}{9}$.

I) Doing $\cos x = y$ and remembering that $\cos 2x = 2\cos^2 x - 1$:

II) The numerator $2y^2 - y - 1 = (2y + 1)(y - 1)$ cancels out in $-\frac{1}{2}$ and $1$; the denominator, in $\pm\frac{\sqrt{2}}{2}$. Sign board:

The quotient is $\ge 0$ to $y < -\frac{\sqrt{2}}{2}$ or $-\frac{1}{2} \le y < \frac{\sqrt{2}}{2}$ or $y \ge 1$.

III) Examining the cycle, for $0 \le x \le 2\pi$:

• $\cos x < -\frac{\sqrt{2}}{2} \iff \frac{3\pi}{4} < x < \frac{5\pi}{4}$;
• $-\frac{1}{2} \le \cos x < \frac{\sqrt{2}}{2} \iff \frac{\pi}{4} < x \le \frac{2\pi}{3}$ or $\frac{4\pi}{3} \le x < \frac{7\pi}{4}$;
• $\cos x \ge 1 \iff x = 0$ or $x = 2\pi$.

Therefore $S = \left\{x \;\middle|\; x = 0 \text{ ou } \frac{\pi}{4} < x \le \frac{2\pi}{3} \text{ ou } \frac{3\pi}{4} < x < \frac{5\pi}{4} \text{ ou } \frac{4\pi}{3} \le x < \frac{7\pi}{4} \text{ ou } x = 2\pi\right\}$.

I) We must have $\dfrac{\cos 2x}{\cos x} \ge 0$.

II) Doing $\cos x = y$: $\dfrac{\cos 2x}{\cos x} \ge 0 \iff \dfrac{2y^2 - 1}{y} \ge 0$.

III) Signal table:

The quotient is $\ge 0$ to $-\frac{\sqrt{2}}{2} \le y < 0$ or $y \ge \frac{\sqrt{2}}{2}$.

IV) Examining the cycle:

• $-\frac{\sqrt{2}}{2} \le \cos x < 0 \iff \frac{\pi}{2} + 2k\pi < x \le \frac{3\pi}{4} + 2k\pi$ or $\frac{5\pi}{4} + 2k\pi \le x < \frac{3\pi}{2} + 2k\pi$;
• $\cos x \ge \frac{\sqrt{2}}{2} \iff 2k\pi \le x \le \frac{\pi}{4} + 2k\pi$ or $\frac{7\pi}{4} + 2k\pi \le x \le 2\pi + 2k\pi$.

Mastery is the meeting of these ranges.

a) If $\cos\alpha = 0$ the equation becomes $-1 = 0$: no solution. If $\cos\alpha \ne 0$, the discriminant is

Then $-\frac{\sqrt{3}}{2} \le \cos\alpha \le \frac{\sqrt{3}}{2}$, $\cos\alpha \ne 0$: $\frac{\pi}{6} \le \alpha \le \frac{5\pi}{6}$, $\alpha \ne \frac{\pi}{2}$.

b) For both roots to be negative: sum $\frac{2}{\cos\alpha} < 0$ and product $\frac{4\cos^2\alpha - 1}{2\cos^2\alpha} > 0$, that is, $\cos\alpha < -\frac{1}{2}$. With the condition of (a): $-\frac{\sqrt{3}}{2} \le \cos\alpha < -\frac{1}{2}$, that is, $\frac{2\pi}{3} < \alpha \le \frac{5\pi}{6}$.

The base must satisfy $0 < \cos x < 1$. Adding the logarithms, the inequality becomes $\log_{\cos x}\big[(1 + 2\cos x)(1 + \cos x)\big] > 1 = \log_{\cos x}\cos x$. As the base is between 0 and 1, the logarithm is decreasing, and the inequality is equivalent to

which never happens (the discriminant of the trinomial is $4 - 8 < 0$). Logo $S = \varnothing$.

The printed template gives $\frac{\pi}{3} < x < \frac{\pi}{2}$ or $\frac{3\pi}{2} < x < \frac{5\pi}{3}$; but, for example, for $\cos x = \frac{1}{4}$ the first member is worth $\log_{1/4}\frac{15}{8} < 0$, and the inequality is false.

$|\operatorname{tg} x| \le 1 \iff -1 \le \operatorname{tg} x \le 1$. The image of $x$ must be at the intersection of the cycle with the angle $r\hat{O}s$ formed by the straight lines $r = OT_1$ ($T_1$ with ordinate 1 on the tangent axis) and $s = OT_2$ ($T_2$ with ordinate $-1$):

which is summarized in $-\frac{\pi}{4} + k\pi \le x \le \frac{\pi}{4} + k\pi$.

Doing $2x = y$, we have $1 \le \operatorname{tg} y < \sqrt{3}$; examining the cycle, $\frac{\pi}{4} + k\pi \le y < \frac{\pi}{3} + k\pi$. Like $x = \frac{y}{2}$:

For $x \in [0, 2\pi]$ $k = 0, 1, 2, 3$ is of interest:

• $k = 0$: $\frac{\pi}{8} \le x < \frac{\pi}{6}$;
• $k = 1$: $\frac{5\pi}{8} \le x < \frac{2\pi}{3}$;
• $k = 2$: $\frac{9\pi}{8} \le x < \frac{7\pi}{6}$;
• $k = 3$: $\frac{13\pi}{8} \le x < \frac{5\pi}{3}$.