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Question: The general value of \(\theta \) satisfying equation \(2{\sin ^2}\theta - 3\sin \theta - 2 = 0\) is ...

The general value of θ\theta satisfying equation 2sin⁡2θ−3sin⁡θ−2=02{\sin ^2}\theta - 3\sin \theta - 2 = 0 is
A. nπ+(−1)nπ6n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{6}
B. nπ+(−1)nπ2n\pi + {\left( { - 1} \right)^n}\dfrac{\pi }{2}
C. nπ+(−1)n5π6n\pi + {\left( { - 1} \right)^n}\dfrac{{5\pi }}{6}
D. nπ+(−1)n+1π6n\pi + {\left( { - 1} \right)^{n + 1}}\dfrac{\pi }{6}

Explanation

Solution

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

“Complete step-by-step answer:”
In the problem, we are given the equation
2sin⁡2θ−3sin⁡θ−2=02{\sin ^2}\theta - 3\sin \theta - 2 = 0 (1)
The above equation is a quadratic equation in sin⁡θ\sin \theta .
First, we need to find the roots of the quadratic equation.
Put x=sin⁡θx = \sin \theta in the above equation, we get
2x2−3x−2=02{x^2} - 3x - 2 = 0 (2)
The above equation is of the form
ax2+bx+c=0a{x^2} + bx + c = 0 (3)
And we know that roots of a quadratic equation of above form is given by
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} (4)
Comparing (2) and (3), we get
(a=2,b=−3,c=−2)( a = 2 , b = - 3, c = - 2) (5)
Using (5) in equation (4), we get
x=3±9+164=3±54=2,−12 ⇒x=2,−12  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⁡θx = \sin \theta , using that now we get,
⇒sin⁡θ=2\Rightarrow \sin \theta = 2 and
⇒sin⁡θ=−12\Rightarrow \sin \theta = - \dfrac{1}{2}
Since the values of the function sin⁡θ\sin \theta lies in between [−1,1]\left[ { - 1,1} \right],root sin⁡θ=2\sin \theta = 2 is not possible and is neglected.
⇒sin⁡θ=−12\Rightarrow \sin \theta = - \dfrac{1}{2} is the solution of equation (1).
We know that sin⁡\sin function is negative in the third and fourth quadrant of the cartesian plane.
Also sin⁡(2π−θ)=−sin⁡(θ)=sin⁡(−θ)\sin \left( {2\pi - \theta } \right) = - \sin \left( \theta \right) = \sin \left( { - \theta } \right)
⇒sin⁡θ=−12=sin⁡(2π−π6)=−sin⁡(π6)=sin⁡(−π6)\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⁡θ=sin⁡α\sin \theta = \sin \alpha is given by
θ=nπ+(−1)nα\theta = n\pi + {\left( { - 1} \right)^n}\alpha
Using equation (6) in above, we get
sin⁡θ=sin⁡(−π6) ⇒α=(−π6) ⇒θ=nπ+(−1)n(−π6) ⇒θ=nπ+(−1)n+1(π6)  \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, θ=nπ+(−1)n+1(π6)\theta = n\pi + {\left( { - 1} \right)^{n + 1}}\left( {\dfrac{\pi }{6}} \right) is the solution of equation (1).
Hence (D). nπ+(−1)n+1π6n\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.