Since the function of the difference between the length of spring and non-compressed length is
[tex]f(\theta)=2cos\theta+\sqrt{3}[/tex]Part A:
If the two lengths are equal then the difference will be 0, then equate f(theta) by 0
[tex]\theta\rightarrow theta[/tex][tex]\begin{gathered} f(\theta)=0 \\ 2cos\theta+\sqrt{3}=0 \end{gathered}[/tex]Subtract root 3 from both sides
[tex]\begin{gathered} 2cos\theta+\sqrt{3}-\sqrt{3}=0-\sqrt{3} \\ 2cos\theta=-\sqrt{3} \end{gathered}[/tex]Divide both sides by 2
[tex]\begin{gathered} \frac{2cos\theta}{2}=-\frac{\sqrt{3}}{2} \\ cos\theta=-\frac{\sqrt{3}}{2} \end{gathered}[/tex]Since the values of cos are negative in the 2nd and 3rd quadrant, then we will find the value of theta on them
[tex]\begin{gathered} \theta=\pi-cos^{-1}(\frac{\sqrt{3}}{2}) \\ \theta=\pi-\frac{\pi}{6}+2\pi n \\ \theta=\frac{5\pi}{6}+2\pi n \end{gathered}[/tex][tex]\begin{gathered} \theta=\pi+\frac{\pi}{6}+2\pi n \\ \theta=\frac{7\pi}{6}+2\pi n \end{gathered}[/tex]n is any integers
Part B:
We will replace theta with 2 theta
[tex]\begin{gathered} 2\theta=\pi-(cos^{-1}\frac{\sqrt{3}}{2}) \\ 2\theta=\pi-\frac{\pi}{6} \\ 2\theta=\frac{5}{6}\pi \\ \frac{2\theta}{2}=\frac{\frac{5\pi}{6}}{2} \\ \theta=\frac{5\pi}{12} \end{gathered}[/tex][tex]\begin{gathered} 2\theta=\pi+cos^{-1}(\frac{\sqrt{3}}{2}) \\ 2\theta=\pi+\frac{\pi}{6} \\ 2\theta=\frac{7\pi}{6} \\ \frac{2\theta}{2}=\frac{\frac{7\pi}{6}}{2} \\ \theta=\frac{7\pi}{12} \end{gathered}[/tex]n = 1 because the interval from [0, 2pi]
Then the value of theta in part 2 is half the value of theta in part 1
The function with 2theta is
[tex]f(2\theta)=2cos2\theta+\sqrt{3}[/tex]Part C:
The other given equation is
[tex]g(x)=1-sin^2\theta+\sqrt{3}[/tex]We will equate the two functions to find the time
[tex]\begin{gathered} f(\theta)=g(\theta) \\ 2cos\theta+\sqrt{3}=1-sin^2\theta+\sqrt{3} \end{gathered}[/tex]Subtract root 3 from both sides
[tex]\begin{gathered} 2cos\theta+\sqrt{3}-\sqrt{3}=1-sin^2\theta+\sqrt{3}-\sqrt{3} \\ 2cos\theta=1-sin^2\theta \end{gathered}[/tex]Since
[tex]1-sin^2\theta=cos^2\theta[/tex]Then
[tex]2cos\theta=cos^2\theta[/tex]Subtract 2cos theta from both sides
[tex]\begin{gathered} 2cos\theta-2cos\theta=cos^2\theta-2cos\theta \\ 0=cos^2\theta-2cos\theta \end{gathered}[/tex]Take cos theta as a common factor
[tex]\begin{gathered} cos\theta(\frac{cos^2\theta}{cos\theta}-\frac{2cos\theta}{cos\theta})=0 \\ cos\theta(cos\theta-2)=0 \end{gathered}[/tex]Equate each factor by 0 to find the value of theta
[tex]\begin{gathered} cos\theta=0 \\ \theta=0,\theta=\pi \end{gathered}[/tex][tex]\begin{gathered} cos\theta-2=0 \\ cos\theta-2+2=0+2 \\ cos\theta=2\rightarrow neglect\text{ it because 0}\leq cos\theta\leq1 \end{gathered}[/tex]Then they have the same length at the time
[tex]\theta=0,\theta=\pi[/tex]