Question #311454

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).


1
Expert's answer
2022-03-18T17:16:31-0400

a)

Let T the main period of the function y=sin3xy=sin3x, let f(x)=sin3xf(x)=sin3x, then

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

For T to be the period of the function, must be: sin(3x+3T)=sin3xsin(3x+3T)=sin3x

Because the period of sinxsinx is 2π2\pi, then 3T=2π    T=2π33T=2\pi\implies T=\frac{2\pi}3

Answer: T=2π3T=\frac{2\pi}3


b)

4sin3x=3    sin3x=34    3x=arcsin34,4sin3x=3\implies sin3x=\frac34\implies 3x=arcsin\frac34, and the second solution is 3x=πarcsin343x=\pi-arcsin\frac34

Make up a system and add a period 2π2\pi :

{3x=arcsin34+2πn3x=πarcsin34+2πn    \begin{cases} 3x=arcsin\frac34 +2\pi n\\ 3x= \pi-arcsin\frac34 + 2\pi n \end{cases}\implies {x=arcsin343+2πn3x=πarcsin343+2πn3\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} nn is integer

for x[0;2π]:x\isin[0;2\pi]:

n=0    x=arcsin343,x=πarcsin343n=0\implies x=\frac{arcsin\frac34}3,x=\frac{\pi - arcsin\frac34}3

n=1    x=arcsin343+2π3,x=πarcsin343+2π3n=1\implies x=\frac{arcsin\frac34}3+\frac{2\pi}3,x=\frac{\pi -arcsin\frac34}3+\frac{2\pi}3

n=2    x=arcsin343+4π3,x=πarcsin343+4π3n=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;2π][0;2\pi]


c)

To make a graph y=4sin3xy=4sin3x from a graph y=sinxy=sinx, it is necessary to first increase the maximum value of sinxsinx by 4 times and build y=4sinxy=4sinx, then reduce the period of the function by 3 times or narrow along OY and build y=4sin3xy=4sin3x


On picture:

green - y=sinxy=sinx

blue - y=4sinxy=4sinx

red - y=4sin3xy=4sin3x


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