Jtdylktuy

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

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!

$f(x)=0$ 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 $npi + arctan(1/3)$ where $n$ is an integer is the complete set of solutions.

HINT

We have that

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

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

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

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

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

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

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