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

The general value of θ\theta satisfying the equation 2sin⁡2θ−3sin⁡θ−2=02{{\sin }^{2}}\theta -3\sin \theta -2=0 is
A. nπ+(−1)nπ6n\pi +{{(-1)}^{n}}\dfrac{\pi }{6}
В. nπ+(−1)nπ2n\pi +{{(-1)}^{n}}\dfrac{\pi }{2}
C. nπ+(−1)n5π6n\pi +{{(-1)}^{n}}\dfrac{5\pi }{6}
D. nπ+(−1)n7π6n\pi +{{(-1)}^{n}}\dfrac{7\pi }{6}

Explanation

Solution

To solve this question, we need to use the angle sum property of a triangle according to which the sum of all the angles of a triangle is π\pi . A trigonometric equation typically has several solutions or an infinite number of solutions because all trigonometric ratios are periodic. Then, using trigonometric tan techniques, we will simplify the given equation to obtain the final equation.

Formula Used : The trigonometric formulas are:
sin⁡(−θ)=−sin⁡θ\sin (-\theta )=-\sin \theta

Complete step-by-step solution: A trigonometric equation will also have a generic solution, which is stated in a generalized form in terms of "n" and contains all the values that would fulfill the given equation. As per the sin⁡θ\sin \theta formula, the sine of an angle θ\theta , in a right-angled triangle is equal to the ratio of the opposite side and hypotenuse.
The mostly applied value of θ\theta satisfying both the equations
sin⁡θ=12\sin \theta =\dfrac{1}{2}
We will take the angle sum property of the triangle and derive the equation of the angle. Hence, the most general value of θ satisfying the equations sin⁡θ=sin⁡α\sin \theta \text{=}\sin \alpha and cosθ=cosα\text{cos}\theta \text{=cos}\alpha is θ=nπ+α\theta =n\pi +\alpha .
2sin⁡2θ−3sin⁡θ−2=0 ⇒(2sin⁡θ+1)(sin⁡θ−2)=0 ⇒∣sin⁡θ=sin⁡(−π6)∣ ⇒θ=nπ+(−1)n(−π6) \begin{aligned} & 2{{\sin }^{2}}\theta -3\sin \theta -2=0 \\\ & \Rightarrow (2\sin \theta +1)(\sin \theta -2)=0 \\\ & \Rightarrow \left| \sin \theta =\sin \left( \dfrac{-\pi }{6} \right) \right| \\\ & \Rightarrow \theta =n\pi +{{(-1)}^{n}}\left( \dfrac{-\pi }{6} \right) \\\ \end{aligned}
⇒θ=nπ+(−1)n+1(π6) ⇒θ=nπ+(−1)n(7π6) \begin{aligned} & \Rightarrow \theta =n\pi +{{(-1)}^{n+1}}\left( \dfrac{\pi }{6} \right) \\\ & \Rightarrow \theta =n\pi +{{(-1)}^{n}}\left( \dfrac{7\pi }{6} \right) \\\ \end{aligned}

Option ‘D’ is correct

Note: The sine of the angle divided by the cosine of that angle is known as the trigonometric ratio. It can be characterized as the proportion of the perpendicular side to the neighboring side of a right-angled triangle's sides. A solution to an ‘n-order’ ordinary differential equation that uses precisely ‘n’ necessary arbitrary constants.