How do you solve the equation on the interval $ \left[ {0,\,2\pi } \right] $ for $ {\left( {\sin (t)} \right)^2} = \dfrac{1}{2}? $

Hint: To find the solution of the given equation, first of all solve the equation for $ \sin t $ then you will get positive and negative values, and to further solve it you have to know that sine function has positive values in first and second quadrant whereas negative values in third and fourth quadrant. So find the required solutions accordingly.

Complete step-by-step answer:
In order to solve the given equation $ {\left( {\sin (t)} \right)^2} = \dfrac{1}{2} $ we will first simplify or solve the equation for $ \sin t $ as follows
 $
   \Rightarrow {\left( {\sin (t)} \right)^2} = \dfrac{1}{2} \\
   \Rightarrow \sin (t) = \pm \dfrac{1}{{\sqrt 2 }} \\
  $
Now, we have to solve this equation for the value of $ t $ in the interval $ \left[ {0,\,2\pi } \right] $ ,
So we will take inverse of sine function in both sides of the equation
 $
   \Rightarrow {\sin ^{ - 1}}\left( {\sin (t)} \right) = {\sin ^{ - 1}}\left( { \pm \dfrac{1}{{\sqrt 2 }}} \right) \\
   \Rightarrow t = {\sin ^{ - 1}}\left( { \pm \dfrac{1}{{\sqrt 2 }}} \right) \\
  $
Now here we are getting two values, positive and negative so we will solve for both of them separately
For the positive part,
 $ \Rightarrow t = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) $
In the given interval $ \left[ {0,\,2\pi } \right] $ the equation will have two solutions first one in first quadrant and second in the second quadrant, so the solution for this part will be given as
 $
   \Rightarrow t = \dfrac{\pi }{4}\;{\text{and}}\;\pi - \dfrac{\pi }{4} \\
   \Rightarrow t = \dfrac{\pi }{4}\;{\text{and}}\;\dfrac{{3\pi }}{4} \\
  $
Now for the negative part,
 $ \Rightarrow t = {\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) $
In the given interval $ \left[ {0,\,2\pi } \right] $ the equation will have two solutions first one in third quadrant and second in the fourth quadrant, so the solution for this part will be given as
 $
   \Rightarrow t = \pi + \dfrac{\pi }{4}\;{\text{and}}\;2\pi - \dfrac{\pi }{4} \\
   \Rightarrow t = \dfrac{{5\pi }}{4}\;{\text{and}}\;\dfrac{{7\pi }}{4} \\
  $
Therefore $ t = \dfrac{\pi }{4},\;\dfrac{{3\pi }}{4},\;\dfrac{{5\pi }}{4}\;{\text{and}}\;\dfrac{{7\pi }}{4} $ are the solutions for the given equation in the interval $ \left[ {0,\,2\pi } \right] $

Note: Trigonometric functions give positive and negative values in different quadrants according to the function, as we have seen in this question sine function has positive values in first and second quadrant whereas it has negative values in third and fourth quadrant.