Must be done by hand and in radians. problem is in image.

Since both cos(t) and sin^2(t) are symmetrical about t=pi, we can expect symmetrical roots about t=pi.

Also, by inspection, cos(pi)=-1, and sin(pi)=0, we see that t=pi is a solution.

Rewrite the equation as:
1+cos(t)-2sin^2(t)=0
and using sin^2(t)=1-cos^2(t)
1+cos(t)-2(1-cos^2(t))=0
which simplifies to
2cos^2(t)+cos(t)-1=0
set C=cos(t)
2C^2+C-1=0
factor
(C+1)(2C-1)
=>
cos(t)=-1, or cos(t)=1/2
A. cos(t)=-1 is only at t=pi (between 0 and 2pi)
B. cos(t)=1/2 is at Q1 and Q4, thus t=pi/3 and pi=2pi/3

Summary:
t={pi/3, pi, 5pi/3}

1. 8 or -82. 14 or -143. -11, -3, 0, 7, 10