Question

((sqr 3)sin t)-((sqr 1)+(sin^2)*t)=0

Answer 1

Hard to read.
I think you mean

√3sint - √(1+sin^2t) = 0

if so, then ...

√3sint = √(1+sin^2t)
square both sides
3sin^t = 1 + sin^2t
2sin^2 t = 1
sin^2t = 1/2
sint = ±1/√2

which, in degrees would be
x = 45°, 135°, 225° and 315°

BUT, since we squared, all answers should be verified in the equation before we squared.
All answers provide a value of √3/√2 for the RS

but only x= 45 and x=135 give a +√3/√2 for the left side,
x = 225 and x = 315° yiels -√3/√2

so x = 45° or x = 135°
in radians, x = π/4 and 3π/4