Answer :
Let Δx be an increase in the variable such that 0.9x increases by 2π. Then:
[tex]\begin{gathered} 0.9(x+\Delta x)-0.9x=2\pi \\ \Rightarrow0.9x+0.9\Delta x-0.9x=2\pi \\ \Rightarrow0.9\Delta x=2\pi \\ \Rightarrow\Delta x=\frac{2\pi}{0.9} \\ \therefore\Delta x=\frac{20\pi}{9} \end{gathered}[/tex]The period of a function f is a quantity T such that for every x, then:
[tex]f(x)=f(x+T)[/tex]The cosine function has a period of 2π over its argument. In this case, we know that:
[tex]f(x)=\cos (0.9x)[/tex]The argument of cosine is 0.9x and we already know that for an increase of 20π/9 in x, there is an increase of 2π in 0.9x. Therefore:
[tex]\begin{gathered} f(x+\frac{20\pi}{9})=\cos (0.9(x+\frac{20\pi}{9})) \\ =\cos (0.9x+0.9\cdot\frac{20\pi}{9}) \\ =\cos (0.9x+2\pi) \\ =\cos (0.9x) \\ =f(x) \end{gathered}[/tex]Then, for every x we know that:
[tex]f(x+\frac{20\pi}{9})=f(x)[/tex]Therefore, the period of f is:
[tex]\frac{20\pi}{9}[/tex]Following a similar process, we can find the period of the function g. Since we know that the period of the sine function is also 2π.
[tex]\begin{gathered} 5\pi\Delta x=2\pi \\ \Rightarrow\Delta x=\frac{2\pi}{5\pi} \\ \therefore\Delta x=\frac{2}{5} \end{gathered}[/tex]Therefore, the period of g is:
[tex]\frac{2}{5}[/tex]