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Question

The solution of the equation cos2 x+sin x+1=0 lies in the interval
(a) -π/4, π/4
(b) π/4, 3π/4
(c) 3π/4, 5π/4
(d) 5π/4, 7π/4

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Solution

Given equation:
cos2x + sinx + 1 = 0(1 - sin2x) + sinx + 1 = 02 - sin2x + sinx = 0 sin2x - sinx - 2 = 0 sin2x -2 sinx + sinx - 2 = 0 sinx ( sinx - 2 ) + 1 ( sinx - 2 ) = 0 (sinx - 2) ( sinx + 1) = 0
sinx - 2 = 0 or sinx + 1 =0
sinx = 2 or sin x =-1
Now, sinx = 2 is not possible.
And,
sin x = -1 sin x = sin 3π2 x = nπ+-1n3π2
For n = 0, x = 3π2, for n = 1, ​x = 7π2 and so on.
Hence, 3π2 lies in the interval 5π4, 7π4.

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