Verified answer

F = 2^log [cos 4x – cos 2x + (k+3)]

ln F = log [cos 4x – cos 2x + (k+3)] ln 2

derivative

[1/F] dF/dx = ln 2 {1/[cos 4x – cos 2x + (k+3)] } {- 4 sin 4x + 2 sin 2x}

dF/dx = F ln 2 {1/ [cos 4x – cos 2x + (k+3)] } {- 4 sin 4x + 2 sin 2x} = 0

gives

sin 2x = 2 sin 4x

sin 2x = 2 * 2 sin 2x cos 2x

sin 2x[1 - 4 cos 2x] = 0

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either sin 2x = 0 .............. x = 0 ........... F = 2^log [k+3] = 4 =2^2

log [k+3] =2

solve yourself from here??????

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or

1 - 4 cos 2x = 0

cos 2x = 1/4

cos 4x = 2 cos^2 (2x) - 1 = 1/8 - 1 = - 7/8

F = 2^log [- 7/8 - 1/4 + k+3] = 4

F = 2^log [- 9/8 + k+3] = 2^2

log [15/8 + k] = 2

solve yourself from here??????

1) find dF/dx

2) solve dF/dx = 0 : find the value(s) of x that make dF/dx zero

3) plug each value found in (2) into the expression F(x) = 4, then solve for k.