In the interactive book: complete chapter VIII, with all demonstrations and exercises.

Higher than a line

Question $\operatorname{sen} x>1/2$ asks for all positions where the cab is above height line $1/2$. In one turn, the cab crosses this line at $30°$ and $150°$. Between them, it passes through $90°$, where the height is 1. Therefore, in $[0°,360°)$:

$$30°An inequality is a condition with inequality. Your solution is usually a range or a union of ranges. The notation $(a,b)$ excludes the ends; $[a,b]$ includes both; $[a,b)$ includes only the first.

Demonstration by drawing

In the upper semicircle, the heights rise from 0 to 1 and then fall to 0. The horizontal height line $1/2$ cuts this path at two points. The arc between them is above the line; the rest is below. This justifies choosing the interval between crossings.

If the condition was $\operatorname{sen} x\ge1/2$, we would include both crossovers: $[30°,150°]$. For $\operatorname{sen} x<1/2$, in the range $[0°,360°)$, we would have $[0°,30°)\cup(150°,360°)$. The symbol $\cup$ means bringing together the two sets.

A recipe that also works for cosine

First choose the study interval. Then find the points of equality. Draw the horizontal line for sine or vertical line for cosine. Choose the bows that satisfy the comparison and check that the ends fit.

For example, $\cos x>1/2$ asks for points to the right of the vertical line $x=1/2$ of the plane. To avoid confusing the coordinate with the angle, think of “horizontal position greater than $1/2$”. In the angle range $[0°,360°)$, the answer is:

$$[0°,60°)\cup(300°,360°).$$

On the entire real line, the same solution can be written as $-60°+360°k

Tangent: beware of interruptions

In each interval $(-90°+180°k,90°+180°k)$, the tangent grows from unbounded negative values ​​to unbounded positive values. It does not exist at the ends. Like this:

$$\tan x>1\iff45°+180°kAt $[0°,360°)$, this gives $(45°,90°)\cup(225°,270°)$. For $\tan x<1$, in the same range, the result is $[0°,45°)\cup(90°,225°)\cup(270°,360°)$.

Do not multiply an inequality by $\cos x$ without knowing its sign. Multiplying by a negative number reverses the direction; multiplying by zero can destroy equivalence. Separating intervals by sign avoids this error.

Before calculating, look at the limits

Like $-1\le\operatorname{sen} x\le1$, condition $\operatorname{sen} x>2$ never happens. $\operatorname{sen} x<2$ is valid for all reais. For $\cos x\ge1$, only angles of full turns are suitable: $x=360°k$.

Playing with the disc

Paint the arcs with positive sine blue and those with negative cosine with stripes. The part that receives both marks is the second quadrant: $90°intersection, that is, the common part.

In everyday life

  • Time at the top. How much time during each turn the cabin spends above a certain height is a sine inequality.
  • Solar panel. A panel only generates useful energy while the Sun is above a certain angle: an interval of hours, solution of an inequality.
  • Security cameras. An object appears in the image when the angle with the camera axis is less than half the aperture — a cosine inequality.
Field of view of a camera: a cosine inequality decides what appears in the image.

Exercises

Try solving it in your notebook before opening the solution. Exercises 1 and 2 are input; 3 and 4, for practice; 5 and 6, for deepening.

1. Solve $\operatorname{sen} x\le0$ into $[0°,360°)$.

See resolution 1

$\{0°\}\cup[180°,360°)$. The leading zero also has sine zero and should appear.

2. Solve $\cos x<0$ into $[0°,360°)$.

See resolution 2

$(90°,270°)$, the left semicircle without the endpoints.

3. Solve $\operatorname{sen} x\ge1/2$ into $[0°,360°)$.

See resolution 3

$[30°,150°]$. Equality includes the ends.

4. Solve $\tan x\ge0$ into $[0°,180°)$.

See resolution 4

$[0°,90°)$. At $90°$ the tangent does not exist and is then negative until $180°$, which is outside the range.

5. Solve $\cos x>3/2$ to real $x$.

See resolution 5

There is no solution: the cosine never exceeds 1.

6. Simultaneously solve $\operatorname{sen} x>0$ and $\cos x>0$ in $[0°,360°)$.

See resolution 6

The first condition gives $(0°,180°)$ and the second $[0°,90°)\cup(270°,360°)$. The common part is $(0°,90°)$.

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