The general values of - satisfying the equation 2sin2- -3 sin- -2-0 is-

The correct option is D nπ+( 1)^n7π/6 On solving the quadratic equation, 2sin^2θ 3 sinθ 2=0 nbsp; (sinθ 2)(2sinθ+1)=0 Only possible value : ...

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Standard XII

Mathematics

General Solution of sin theta = sin alpha

The general v...

Question

The general values of $θ$ satisfying the equation $2 s i n^{2} θ - 3 s i n θ - 2 = 0$ is-

n π + (- 1)^{n} π / 6

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n π + (- 1)^{n} π / 2

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n π + (- 1)^{n} 5 π / 6

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n π + (- 1)^{n} 7 π / 6

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Solution

The correct option is D $n π + (- 1)^{n} 7 π / 6$
On solving the quadratic equation,
$2 s i n^{2} θ - 3 s i n θ - 2 = 0$
$(s i n θ - 2) (2 s i n θ + 1) = 0$
Only possible value : $s i n θ = - \frac{1}{2}$
Hence , $θ = n π + (- 1)^{n} 7 π / 6$
Option D

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The general values of θ satisfying the equation 2sin2θ−3sinθ−2=0 is-

The correct option is D nπ+(−1)n7π/6On solving the quadratic equation,2sin2θ−3sinθ−2=0 (sinθ−2)(2sinθ+1)=0Only possible value : sinθ=−12Hence , θ=nπ+(−1)n7π/6Option D

The general values of $θ$ satisfying the equation $2 s i n^{2} θ - 3 s i n θ - 2 = 0$ is-

The correct option is D $n π + (- 1)^{n} 7 π / 6$
On solving the quadratic equation,
$2 s i n^{2} θ - 3 s i n θ - 2 = 0$
$(s i n θ - 2) (2 s i n θ + 1) = 0$
Only possible value : $s i n θ = - \frac{1}{2}$
Hence , $θ = n π + (- 1)^{n} 7 π / 6$
Option D