Since you knew how to solve the first part then I'll just go into the theta thing (:
the Equation you'll use is => sin theta [ squareroot52 sin (theta + 0.588)] =0 from part (c)
so sin theta = 0
we know that sin theta is 0 , at pie and 2pie .... only 0 is accepted => check the range
squareroot52 sin (theta + 0.588) = 0
theta + 0.588 = 0 , pie , 2 pie (cuz I told you ,we know that sin theta is ZERO at these three angles )
so theta = pie - 0.588 = 2.553
theta =2pie -0.588 = 5.70 { rejected TOO big not within the range }
therefore
Theta = 0 rad , 2.553 rad This is one of the Tricks my teacher uses , I hope you got it (:
[I might have to go any minute cuz my bro's got a project -.- ]