Part A)
Since the function f represents the difference in length of the spring from its non-compressed position, then, when the spring is not compressed, the value of f(θ) is equal to 0:
[tex]f(\theta)=0[/tex]Replace the expression for f(θ) and solve for θ:
[tex]\begin{gathered} 2\cos (\theta)+\sqrt[]{3}=0 \\ \Rightarrow2\cos (\theta)=-\sqrt[]{3} \\ \Rightarrow\cos (\theta)=-\frac{\sqrt[]{3}}{2} \\ \Rightarrow\theta=\arccos (-\frac{\sqrt[]{3}}{2}) \\ \\ \therefore\theta=\pm\frac{5}{6}\pi+2\pi k;k\in\Z \end{gathered}[/tex]Part B)
Replace θ=2θ and find the solutions in the given interval:
[tex]\begin{gathered} \Rightarrow2\theta=\pm\frac{5}{6}\pi+2\pi k \\ \Rightarrow\theta=\pm\frac{5}{12}\pi+\pi k \end{gathered}[/tex]For k=0, the positive solution lies in the interval [0,2π):
[tex]\theta_1=\frac{5}{12}\pi[/tex]For k=1, both solutions lie in the interval [0,2π):
[tex]\begin{gathered} \theta_2=-\frac{5}{12}\pi+\pi=\frac{7}{12}\pi \\ \theta_3=\frac{5}{12}\pi+\pi=\frac{17}{12}\pi \end{gathered}[/tex]For f=2, the negative solution lies in the interval [0,2π):
[tex]\theta_4=-\frac{5}{12}\pi+2\pi=\frac{19}{12}\pi[/tex]Then, the solutions in the interval [0,2π) are:
[tex]\mleft\lbrace\frac{5}{12}\pi,\frac{7}{12}\pi,\frac{17}{12}\pi,\frac{19}{12}\pi\mright\rbrace[/tex]