Solving equation with $\cos$ and $\sin$

Question

$f(x) = 3\cos(x) - 9\sin(x)$

Is there an easy way to solve $f(x) = 0$?

I'm drawing a blank. It seems impossible and the solution to the question I'm trying to do skips over showing the solving.

Thanks!

Answer 1

HINT

We have that

$$f(x)=3\cos x - 9 \sin x =0 \implies 3\cos x \cdot (1-3 \tan x)=0$$

and $\cos x=0$ is not a solution.

Answer 2

$f(x)=0$ means $3 \cos x = 9 \sin x$.

Now if $\cos x=0$, then $\sin x=1$ or $-1$, so values of $x$ where $\cos x=0$ are not a solution to the equation. Divide by $9 \cos x$ on both sides,

$\tan x = 1/3$

Now you can probably compute $\arctan(1/3)$ by a calculator and $n\pi + \arctan(1/3)$ where $n$ is an integer is the complete set of solutions.

Answer 3

$$f(x) = 3\cos x-9\sin x$$

$$f(x) = 0 \implies 0 = 3\cos x-9\sin x$$

$$\implies 9\sin x = 3\cos x \implies \tan x = \frac{1}{3}$$

$\tan x$ takes positive values in the first and third quadrants.

For the first quadrant,

$$x = \arctan \frac{1}{3}$$

For the third quadrant,

$$x = \pi+\arctan \frac{1}{3}$$

That is the solution for $x \in [0, 2\pi]$. For a general solution, $\tan x$ is periodic every $\pi$ radians, so for all $n \in \mathbb{Z}$,

$$x = \pi n+\arctan \frac{1}{3}$$