The given function for the difference in length is presented as follows;
[tex]f( \theta) = 2 \cdot cos( \theta) +  \sqrt{3} [/tex]
When the pogo stick will be equal to its non compressed length, the difference is zero, therefore;
[tex]f( \theta) = 2 \cdot cos( \theta) +  \sqrt{3}  = 0[/tex]
[tex] 2 \cdot cos( \theta)  =  -  \sqrt{3} [/tex]
[tex]\theta=  arccos \left(  \frac{ -  \sqrt{3} }{2}  \right)   [/tex]
Which gives;
[tex] \theta =  \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]
[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{6} [/tex]
Part B; If the angle was doubled, we have;
[tex]f( \theta) = 2 \cdot cos(2 \cdot \theta) +  \sqrt{3}  = 0[/tex]
Therefore;
[tex] 2 \cdot cos(2 \cdot \theta)  =  -  \sqrt{3} [/tex]
Which gives;
[tex] \theta =  \frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]
[tex] \theta = -\frac{12 \cdot \pi \cdot n1 + 5\cdot \pi}{12} [/tex]
Between 0 and 2•π, we have;
[tex] \theta =  \frac{5\cdot \pi}{12} [/tex]
[tex] \theta = \frac{ 17\cdot \pi}{12} [/tex]
Part C;
The toddler's pogo stick is presented as follows;
[tex]g( \theta) = 1- son^2( \theta) +  \sqrt{3} [/tex]
Integrating the original function between 0 and theta gives;
[tex]2 \cdot sin( \theta) +  \sqrt{3}\cdot  \theta [/tex]
The original length = 
Therefore, when the lengths are equal, we have;
[tex]1- son^2( \theta) +  \sqrt{3} =   2 \cdot sin( \theta) +  \sqrt{3}\cdot  \theta [/tex]