A variable star is one whose brightness alternately increases and decr

Step 1
Start with a sine function, that has average value 0, magnitude ${\displaystyle \pm 1}$, period ${\displaystyle 2\pi }$
${\displaystyle f\left(t\right)=\sin }$
To make the period 5.4 days, multiply t by ${\displaystyle \frac{2\pi }{5.4}}$
${\displaystyle f\left(t\right)=\sin \left(\frac{2\pi }{5.4}t\right)}$
Now whenever t is a multiple of 5.4, it will be the sine of an integer multiple of ${\displaystyle 2\pi }$
Step 2
The make the average value (the center horizontal line that the sine function moves back and forth over) 4.0, add to the function 4.0, It shifts the graph up by 4 units
${\displaystyle f\left(t\right)=4+\sin \left(\frac{2\pi }{5.4}t\right)}$
Step 3
To change the magnitude to ${\displaystyle \pm 0.35}$ multiply the sine by 0.35. Since the sine bounces back and forth between -1 and 1, this makes it so it bounces back and forth between -0.35 and 0.35
${\displaystyle f\left(t\right)=4+0.35\sin \left(\frac{2\pi }{5.4}t\right)}$
Since they don't give any information about what the brightness should be at a specific time, we don't need to add or subtract anything from t to horizontally shift the graph.

Step 1
NSK
There are many functions that alternately increase and decrease (called periodic functions), but the simplest of these is the sine function,
$y=A\sin (\frac{2\pi }{T})t-\varphi )+b$
Here y and t are the variables: magnitude and time in your case.
A is called the amplitude; it tells us the maximum variation of y. In your case, A = 0.35.
T is called the period; it tells us the duration of one cycle. In your case, T = 5.4 days.
b is the off-set in the y-direction, or average value of y. In your case, b = 4.4.
$\varphi$ is called the phase shift; it tells us by how much the initial value of y is off-set from the average value, y=b. In your case
$\varphi =0$
Putting it all in, you get
$y=0.35\sin ((\frac{2\pi }{5.4})t)+4.4$