Answer to Question #311454 in Trigonometry for bookaddict

Question

Consider 𝑔(𝑥)=4sin3𝑥

a. Write down the period of 𝑔(𝑥)

b. Write down the number of solutions to the equation 𝑔(𝑥)=3, for 0≤𝑥≤2𝜋

c. Starting with the graph of 𝑦=sin𝑥, state the transformations which can be used to sketch 𝑔(𝑥).

Answer 1

a)

Let T the main period of the function $y=sin3x$, let $f(x)=sin3x$, then

$f(x +T)=sin3(x+T)=sin(3x+3T)$

For T to be the period of the function, must be: $sin(3x+3T)=sin3x$

Because the period of $sinx$ is $2\pi$, then $3T=2\pi\implies T=\frac{2\pi}3$

Answer: $T=\frac{2\pi}3$

b)

$4sin3x=3\implies sin3x=\frac34\implies 3x=arcsin\frac34,$ and the second solution is $3x=\pi-arcsin\frac34$

Make up a system and add a period $2\pi$ :

$\begin{cases} 3x=arcsin\frac34 +2\pi n\\ 3x= \pi-arcsin\frac34 + 2\pi n \end{cases}\implies$ $\begin{cases} x=\frac{arcsin\frac34}3+\frac{2\pi n}3\\ x=\frac{\pi-arcsin\frac34}3+\frac{2\pi n}3 \end{cases}$ $n$ is integer

for $x\isin[0;2\pi]:$

$n=0\implies x=\frac{arcsin\frac34}3,x=\frac{\pi - arcsin\frac34}3$

$n=1\implies x=\frac{arcsin\frac34}3+\frac{2\pi}3,x=\frac{\pi -arcsin\frac34}3+\frac{2\pi}3$

$n=2\implies x=\frac{arcsin\frac34}3+\frac{4\pi}3,x=\frac{\pi-arcsin\frac34}3+\frac{4\pi}3$

Answer: 6 solutions in $[0;2\pi]$

c)

To make a graph $y=4sin3x$ from a graph $y=sinx$, it is necessary to first increase the maximum value of $sinx$ by 4 times and build $y=4sinx$, then reduce the period of the function by 3 times or narrow along OY and build $y=4sin3x$

On picture:

green - $y=sinx$

blue - $y=4sinx$

red - $y=4sin3x$