Question

The general value of $\theta $ satisfying equation $2{\sin ^2}\theta - 3\sin \theta - 2 = 0$ is
A. $n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{6}$
B. $n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{2}$
C. $n\pi + {\left( { - 1} \right)^n}\dfrac{{5\pi }}{6}$
D. $n\pi + {\left( { - 1} \right)^{n + 1}}\dfrac{\pi }{6}$

Answer 1

Hint: The equation is a quadratic equation in $\sin \theta $.The roots are needed to be calculated using properties of quadratic equation and then the general trigonometric solution for $\sin \theta = \sin \alpha $ is needed to be used.

“Complete step-by-step answer:”
In the problem, we are given the equation
$2{\sin ^2}\theta - 3\sin \theta - 2 = 0$ (1)
The above equation is a quadratic equation in $\sin \theta $.
First, we need to find the roots of the quadratic equation.
Put $x = \sin \theta $ in the above equation, we get
$2{x^2} - 3x - 2 = 0$ (2)
The above equation is of the form
$a{x^2} + bx + c = 0$ (3)
And we know that roots of a quadratic equation of above form is given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ (4)
Comparing (2) and (3), we get
$( a = 2 , b = - 3, c = - 2)$ (5)
Using (5) in equation (4), we get
$
  x = \dfrac{{3 \pm \sqrt {9 + 16} }}{4} = \dfrac{{3 \pm 5}}{4} = 2, - \dfrac{1}{2} \\
   \Rightarrow x = 2, - \dfrac{1}{2} \\
$
Since in equation (2) we had put $x = \sin \theta $, using that now we get,
$ \Rightarrow \sin \theta = 2$ and
$ \Rightarrow \sin \theta = - \dfrac{1}{2}$
Since the values of the function $\sin \theta $ lies in between $\left[ { - 1,1} \right]$,root $\sin \theta = 2$ is not possible and is neglected.
$ \Rightarrow \sin \theta = - \dfrac{1}{2}$ is the solution of equation (1).
We know that $\sin $ function is negative in the third and fourth quadrant of the cartesian plane.
Also $\sin \left( {2\pi - \theta } \right) = - \sin \left( \theta \right) = \sin \left( { - \theta } \right)$
$ \Rightarrow \sin \theta = - \dfrac{1}{2} = \sin \left( {2\pi - \dfrac{\pi }{6}} \right) = - \sin \left( {\dfrac{\pi }{6}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)$ (6)

We know that the general solution of the equation $\sin \theta = \sin \alpha $ is given by
$\theta = n\pi + {\left( { - 1} \right)^n}\alpha $
Using equation (6) in above, we get
$
  \sin \theta = \sin \left( { - \dfrac{\pi }{6}} \right) \\
   \Rightarrow \alpha = \left( { - \dfrac{\pi }{6}} \right) \\
   \Rightarrow \theta = n\pi + {\left( { - 1} \right)^n}\left( { - \dfrac{\pi }{6}} \right) \\
   \Rightarrow \theta = n\pi + {\left( { - 1} \right)^{n + 1}}\left( {\dfrac{\pi }{6}} \right) \\
$

Therefore, $\theta = n\pi + {\left( { - 1} \right)^{n + 1}}\left( {\dfrac{\pi }{6}} \right)$ is the solution of equation (1).
Hence (D). $n\pi + {\left( { - 1} \right)^{n + 1}}\dfrac{\pi }{6}$ is the correct answer.
Note: It is advised to remember the general solutions of trigonometric equations in problems like above. Also, the sign of the trigonometric ratio in a quadrant is important while finding the value of angle in a trigonometric equation.