How do you solve $ 2{\sin ^2}x + 3\sin x + 1 = 0 $ and find all the solutions in the interval $ [0,2\pi ) $ ?

Hint: On replacing $ \sin x $ in the given equation with any other variable, we see that the equation is a polynomial equation. Now, the degree of a polynomial equation is the highest power of the variable used in the polynomial. A polynomial equation has exactly as many roots as its degree, so the given equation will have two roots that can be found by factoring the equation or by a special formula called completing the square method if we are not able to factorize the equation. So, after finding the possible values of $ \sin x $ , we can find the values of x.

Complete step-by-step answer:
We have to solve the equation $ 2{\sin ^2}x + 3\sin x + 1 = 0 $
Let $ \sin x = t $
 $ \Rightarrow 2{t^2} + 3t + 1 = 0 $
The equation written above is a quadratic polynomial equation, so we can solve the above equation by factorization to find the value of t, as follows –
 $
  2{t^2} + 2t + t + 1 = 0 \\
  2t(t + 1) + 1(t + 1) = 0 \\
   \Rightarrow (2t + 1)(t + 1) = 0 \\
   \Rightarrow 2t + 1 = 0,\,t + 1 = 0 \\
   \Rightarrow t = \dfrac{{ - 1}}{2},\,t = - 1 \;
  $
Now replacing t with the original value, we get –
 $
  \sin x = \dfrac{{ - 1}}{2},\,\sin x = - 1 \\
   \Rightarrow \sin x = - \sin \dfrac{\pi }{6},\,\sin x = - \sin \dfrac{\pi }{2}\,\,\,\,(\sin \dfrac{\pi }{6} = \dfrac{1}{2},\,\sin \dfrac{\pi }{2} = 1) \\
  $
Now,
 $
   - \sin \dfrac{\pi }{6} = \sin (\pi + \dfrac{\pi }{6}),\, - \sin \dfrac{\pi }{6} = \sin (2\pi - \dfrac{\pi }{6}) \\
   \Rightarrow - \sin \dfrac{\pi }{6} = \sin \dfrac{{7\pi }}{6},\, - \sin \dfrac{\pi }{6} = \sin \dfrac{{11\pi }}{6} \;
  $
And
 $
   - \sin \dfrac{\pi }{2} = \sin (\pi + \dfrac{\pi }{2}),\, - \sin \dfrac{\pi }{2} = \sin (2\pi - \dfrac{\pi }{2}) \\
   \Rightarrow - \sin \dfrac{\pi }{2} = \sin \dfrac{{3\pi }}{2} \;
  $
Thus,
 $
  \sin x = \sin \dfrac{{7\pi }}{6} = \sin \dfrac{{11\pi }}{6} = \sin \dfrac{{3\pi }}{2} \\
   \Rightarrow x = \dfrac{{7\pi }}{6} = \dfrac{{11\pi }}{6} = \dfrac{{3\pi }}{2} \;
  $
Hence, the solution of the equation $ 2{\sin ^2}x + 3\sin x + 1 = 0 $ is $ x = \dfrac{{7\pi }}{6} $ or $ \dfrac{{11\pi }}{6} $ or $ \dfrac{{3\pi }}{2} $ .
So, the correct answer is “ $ \dfrac{{7\pi }}{6} $ or $ \dfrac{{11\pi }}{6} $ or $ \dfrac{{3\pi }}{2} $ ”.

Note: The standard form of a quadratic polynomial equation is $ a{x^2} + bx + c = 0 $ .While solving an equation by factorization, the condition to form factors is that we have to express the coefficient of x as a sum of two numbers such that their product is equal to the product of the coefficient of $ {x^2} $ and the constant $ c $ that is $ a \times c = {b_1} \times {b_2} $ . We are given that the solution lies in the interval $ [0,2\pi ) $ that is the value of x can be greater than or equal to zero but smaller than $ 2\pi $ .