find the max.value of sin^2 x - 3sin x + 2

1.	find the max.value of sin^2 x - 3sin x + 2
Answer» 2 sin^2 (x) - 3 sinx + 2. Let y = 2 sin^2 (x) - 3 sinx + 2 For y to be minimum or maximum,dy/dx = 0 Or, 4 sinx cosx - 3 cosx = 0Or, cosx (4 sinx - 3) = 0 cosx = 0, sinx = 3/4x = 90°, 48.59°. d^2 y / dx^2= 4 (cos^2 (x) - sin^2 (x)) + 3 sinx= 4 cos (2x) + 3 sinx When x = 90°, d^2 y / dx^2 = -1 < 0So, y is maximum when x = 90° When x = 48.59°, d^2 y / dx^2 = 1.75 > 0So, y is minimum when x = 48.59° y (max) = 2 sin^2 (90°) - 3 sin (90°) + 2= 1

Answer»

2 sin^2 (x) - 3 sinx + 2.

Let y = 2 sin^2 (x) - 3 sinx + 2

For y to be minimum or maximum,dy/dx = 0

Or, 4 sinx cosx - 3 cosx = 0Or, cosx (4 sinx - 3) = 0

cosx = 0, sinx = 3/4x = 90°, 48.59°.

d^2 y / dx^2= 4 (cos^2 (x) - sin^2 (x)) + 3 sinx= 4 cos (2x) + 3 sinx

When x = 90°, d^2 y / dx^2 = -1 < 0So, y is maximum when x = 90°

When x = 48.59°, d^2 y / dx^2 = 1.75 > 0So, y is minimum when x = 48.59°

y (max) = 2 sin^2 (90°) - 3 sin (90°) + 2= 1

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