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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`

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The correct Answer is:
` alpha in [-(3)/(2),(1)/(2)] and x=(n x )/(2)+((-1)^(n))/(2)sin^(-1)(1-sqrt(2 alpha +3))`
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ARIHANT MATHS ENGLISH-TRIGONOMETRIC EQUATIONS AND INEQUATIONS-Exercise (Subjective Type Questions)
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