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 2π, then 3T=2π⟹T=32π
Answer: T=32π
b)
4sin3x=3⟹sin3x=43⟹3x=arcsin43, and the second solution is 3x=π−arcsin43
Make up a system and add a period 2π :
{3x=arcsin43+2πn3x=π−arcsin43+2πn⟹ {x=3arcsin43+32πnx=3π−arcsin43+32πn n is integer
for x∈[0;2π]:
n=0⟹x=3arcsin43,x=3π−arcsin43
n=1⟹x=3arcsin43+32π,x=3π−arcsin43+32π
n=2⟹x=3arcsin43+34π,x=3π−arcsin43+34π
Answer: 6 solutions in [0;2π]
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
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