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If 2 cos2θ + 11 sinθ - 7 = 0, then the value of sinθ, where [0 ≤ θ ≤ \(\frac{{\rm{\pi }}}{2}\)

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If 2 cos2θ + 11 sinθ - 7 = 0, then the value of sinθ, where [0 ≤ θ ≤ \(\frac{{\rm{\pi }}}{2}\)


1). \( - \frac{1}{2}\)
2). 1/2
3). 5
4). \(\frac{1}{{\sqrt 2 }}\)


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1 answers

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Answered by on | Votes 2 |

2cos2θ + 11 sinθ - 7 = 0

2(1 - sin2θ) + 11 sinθ - 7 = 0

2sin2θ - 11 sinθ + 5 = 0

2sin2 θ - 10 sin θ - sin θ + 5 = 0

2sinθ (sinθ - 5) - 1 (sinθ - 5) = 0

(2sinθ - 1) (sinθ - 5) = 0

∴ sinθ $( = \frac{1}{2})$

[As sinθ = 5 is not possible]

[As we know sinθ = 0 ≤ θ ≤ $(\frac{{\rm{\pi }}}{2})$] 

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