If for all real values of x, 4x^2+1 divided by 64x^2-32x sin alpha+29>1/32 then alpha lies in the interval (ans 0,pi/3)

Dear Sudha ,

The solution can be found by reorganizing the equation as ( 4x² + 1 ) * 32 > 64x² - 32xsin(alpha) + 29

or 128x² + 32 > 64x² - 32xsin(alpha) + 29

or 64x² + 32xsin(alpha) + 3 > 0

The above Identity will hold true for all of x if Determinant is less than 0

because then no solution will exist such that 64x² + 32xsin(alpha) + 3 = 0

i e D ≤ 0

or ( 32sin(alpha) )²  - 4*64*3 ≤ 0

or 32sin²(alpha) - 24 ≤ 0

or 4sin²(alpha) - 3 ≤ 0

or sin²(alpha) ≤ 3/4

this will give that alpha should lie in the Interval ( 0 , ∏/3 ) ( as sin(∏/3) = sq. root(3) / 2)

Puneet

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