Math help, please?

Question

If F(x) = Sin(x) Cos^3(x), prove that minimum value of F(x) be when Sin^2(x) = 1/4?

Please take your time and consult with others, i really need the answer within 24 Hrs. This is for my little brother.

Please see my other math questions.

Thank you very much.

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Answer 1

f(x) = sinx * cos^3x

f'(x) = cosx * cos^3x + sinx(3cos^2x{-sinx}) = 0

cos^4x -3cos^2x sin^2x = 0

cos^2x { cos^2x - 3sin^2x} = 0

One answer is cos^2x = 0 or x = (2n-1)*pi/2

The other answer comes from

cos^2-3sin^2x = 0

cos^2x = 3sin^2x

cos^2x + sin^2x = 1

1-sin^2x - 3sin^2x = 0

1 = 4sin^2x

sin^2x = 1/4

Bozo