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The equation sin^4x+cos^4x+sin2x+alpha=0...

The equation `sin^4x+cos^4x+sin2x+alpha=0` is solvable for `-5/2lt=alphalt=1/2` (b) `-3lt=alpha<1` `-3/2lt=alphalt=1/2` (d) `-1lt=alphalt=1`

A

`-(1)/(2) le alpha (1)/(2)`

B

`-3 le alpha le 1`

C

`-(3)/(2) le alpha le (1)/(2)`

D

`-1 le alpha le 1`

Text Solution

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The correct Answer is:
C
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