Question

Please help! maths functions

Answer 1

1. I'm not sure how you're expected to "read off" where they intersect based on an imprecise hand-drawn graph, but we can still find these intersections exactly.

$2\sin(x) = 3 \cos(x)$

$\dfrac{\sin(x)}{\cos(x)} = \dfrac32$

$\tan(x) = \dfrac32$

$x = \tan^{-1}\left(\dfrac32\right) + 180^\circ n$

where $n$ is an integer.

In the interval [0°, 360°], we have solutions at

$x \approx 56.31^\circ \text{ or } x \approx 236.31^\circ$

From the sketch of the plot, we do see that the intersections are roughly where we expect them to be. (The first is somewhere between 45° and 90°, while the second is somewhere between 225° and 270°.)

2. According to the plot and the solutions from (1), we have

$3 \cos(x) > 2 \sin(x)$

whenever $0^\circ < x < 56.31^\circ$ or $236.31^\circ < x < 360^\circ$.

3. Rewrite the inequality as

$2 \sin(x) - 3 \cos(x) \le 0 \implies 3 \cos(x) \ge 2 \sin(x)$

The answer to (1) tells us where the equality $3\cos(x) = 2\sin(x)$ holds.

The answer to (2) tells us where the strict inequality $3\cos(x)>2\sin(x)$ holds.

Putting these solutions together, we have $2\sin(x) - 3\cos(x) \le 0$ whenever $0^\circ < x \le 56.31^\circ$ or $236.61^\circ \le x < 360^\circ$.