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)`

`(\sin(x))/(\cos(x)) = \frac32`

`\tan(x) = \frac32`

`x = \tan^(-1)\left(\frac32\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`.