If 2 sin^3 ⁡x+ sin ⁡2x cos ⁡x+4 sin ⁡x-4=0 has exactly 3 solutions in the interval (0 , nπ2 ) ,n ∈N, then the roots of the equation x^2+ nx+ (n-3 )=0 belong to :

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Accepted Answer

Correct answer: B. (b) 2 sin^3 ⁡x+2 sin ⁡x cos^2 ⁡x+4 sin ⁡x-4=02 sin^3 ⁡x+2 sin ⁡x (1- sin^2 ⁡x)+4 sin ⁡x-4=06 sin ⁡x-4=0 sin ⁡x= 23 n=5 (in the given interval) x^2+5x+2=0x= -5 172 Required interval (- ,0 )

$If 2 sin^{3} x + sin 2 x cos x + 4 sin x - 4 = 0 has exactly 3 solutions in the interval (0, \frac{nπ}{2}), n \in N, then the roots of the equation x^{2} + nx + (n - 3) = 0 belong to :$

Solution

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