You could use Newton's iterative Method and calculus.
First, write f(t)=t2/2-cos(t)-1. Note that t must be measured in radians (not degrees).
Now differentiate: df/dt=f'(t)=t+sin(t).
Newton's Method is:
tn+1=tn-f(tn)/f'(tn) where tn is a value close to where we would expect to find t, the solution of the equation f(t)=0. Each iteration, t1, t2, etc., should bring us closer to the solution for t.
To get the process started we need t0. To find such a value close to the solution t, we could sketch a graph or we could just substitute an arbitrary value for t0. We can't start with t=0 because f'(0)=0 and we would have zero as a denominator (division by zero is not permitted). Let's start with t=1 (radian).
We need a calculator to find each iteration.
t1=t0-f(t0)/f'(t0)=1-f(1)/f'(1)=1-(0.5-cos(1)-1)/(1+sin(1))=1.56493.
t2=t1-f(t1)/f'(t1)=1.47969, t3=1.47817, t4=1.478170266, t5=1.4781702664, t6=1.47817026643.
Depending on how much accuracy we need the process is continued, but we'll take t=1.47817 to be a fairly accurate solution. However, we're not finished yet, because t2=(-t)2 and cos(t)=cos(-t), which implies that the full solution has to be t=±1.47817, making two solutions for t (what's true for t must also be true for -t).
A graph would show the two solutions approximately at t=-1.5 and t=1.5. We could have used either of these values for t0 and arrived at the solution for t much quicker. For example, t0=1.5, t1=1.47827, t2=1.47817.