SATHEE: Trigonometrical Equations 1 Question 14

Trigonometrical Equations 1 Question 14

14. The general value of $θ$ satisfying the equation $2 \sin^{2} θ - 3 \sin θ - 2 = 0$ , is

(1995,2M)

(a) $n π + (- 1)^{n} \frac{π}{6}$

(b) $n π + (- 1)^{n} \frac{π}{2}$

(d) $n π + (- 1)^{n} \frac{7 π}{6}$

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Answer:

Correct Answer: 14. (8)

Solution:

Given, $2 \sin^{2} θ - 3 \sin θ - 2 = 0$

$\Rightarrow (2 \sin θ + 1) (\sin θ - 2) = 0$

$\begin{array}{ll} \Rightarrow & \sin θ = - 1 / 2 \\ [neglecting \sin θ = 2, as | \sin θ | \leq 1] \end{array}$

$∴ θ = n π + (- 1)^{n} (7 π / 6)$

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Trigonometrical Equations 1 Question 14

14. The general value of θ satisfying the equation 2sin2⁡θ−3sin⁡θ−2=0, is

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14. The general value of $θ$ satisfying the equation $2 \sin^{2} θ - 3 \sin θ - 2 = 0$ , is