Question

maths trig functions
please help

Answer 1

1. According to the plots, the curves intersect when $x=90^\circ$ and $x=360^\circ$.

We can confirm this algebraically.

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

$\sin(x) + \cos(x) = 1$

$\sqrt2 \sin\left(x + 45^\circ\right) = 1$

$\sin\left(x + 45^\circ\right) = \dfrac1{\sqrt2}$

$x + 45^\circ = \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n \text{ or } x + 45^\circ = 180^\circ - \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n$

(where $n$ is an integer)

$x + 45^\circ = 45^\circ + 360^\circ n \text{ or } x + 45^\circ = 135^\circ + 360^\circ n$

$x = 360^\circ n \text{ or } x = 90^\circ + 360^\circ n$

We get the two solutions we found in the interval [0°, 360°] with $n=1$ in the first case, and $n=0$ in the second case.

2. We have $\sin(x)=0$ when $x \in \{0^\circ, \pm 180^\circ, \pm360^\circ, \ldots\}$. For the given plot domain [0°, 360°], this happens when $180^\circ < x < 360^\circ$.

3. The domain for both equations is all real numbers in general, but considering the given plot, you could argue the domains would be [0°, 360°].

$\sin(x)$ is bounded between -1 and 1, so $\sin(x) + 2$ is bounded between -1 + 2 = 1 and 1 + 2 = 3, and its range is [1, 3].

Likewise, $-\cos(x)$ is bounded between -1 and 1, so that $-\cos(x)+3$ is bounded between -1 + 3 = 2 and 1 + 3 = 4, so its range would be [2, 4].