Question

help me to solve this question​

Answer 1

Answer:

hey here is your solution

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Step-by-step explanation:

$so \: here \: given \: quadratic \: equation \: is \\ 2sin {}^{2} x + 2 \sqrt{2} \: sinx - 3 = 0 \\ \\ so \: comparing \: it \: with \: \: a.sin {}^{2} x + b.sinx + c = 0 \\ we \: get \\ a = 2 \: \\ b = 2 \sqrt{2} \\ c = - 3 \\ \\ now \: applying \\ Δ = b {}^{2} - 4ac \\ = (2 \sqrt{2} ) {}^{2} - 4 \times 2 \times ( - 3) \\ = 8 - ( - 24) \\ = 8 + 24 \\ = 32 \\ \\ so \: by \: applying \: formula \: method \\ we \: get \\ \\ sin \: x = \frac{ - b \: ± \: \sqrt{b {}^{2} - 4ac} }{2a} \\ \\ = \frac{ - 2 \sqrt{2} \: ± \: \sqrt{32} }{2 \times 2} \\ \\ = \frac{ - 2 \sqrt{2} \: ± \: 4 \sqrt{2} }{4} \\ \\ sin \: x = \frac{ - 2 \sqrt{2} \: + \: 4 \sqrt{2} }{4} \: \: or \: \: sin \: x = \frac{ - 2 \sqrt{2} - 4 \sqrt{2} }{4} \\ \\ = \frac{2 \sqrt{2} }{4} \: \: \: or \: \: \: = \frac{ - 6 \sqrt{2} }{4} \\ \\ sin \: x= \frac{ \sqrt{2} }{2} \: \: \: or \: \: sin \: x= \frac{ - 3 \sqrt{2} }{2}$