ii)
From MF26, we can obtain the expansion of cosx to be :
1 - x²/2! + x⁴/4! - x⁶/6! + ...
So ln (cos2x) = ln (1 - (2x)²/2! + (2x)⁴/4! - (2x)⁶/6! +...)
= ln (1 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) )
Combine this with the expansion of ln(1 + x) by replacing x with (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)
Then expand up to term in x⁶
So ln (1 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) )
= (-2x² + ⅔x⁴ - 4/45 x⁶ + ...) - (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)²/2 + (-2x² + ⅔x⁴ - 4/45 x⁶ + ...)³/3
= -2x² + ⅔x⁴ - 4/45 x⁶ + ... - ½(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...) + ⅓(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)(-2x² + ⅔x⁴ - 4/45 x⁶ + ...)
= -2x² + ⅔x⁴ - 4/45 x⁶ + ... - ½( -2x²(-2x²) - 2x²(⅔x⁴) + ⅔x⁴(-2x²) + ...) + ⅓((-2x²)(-2x²)(-2x²) + ...)
= -2x² + ⅔x⁴ - 4/45 x⁶ - 2x⁴ + 4/3 x⁶ - 8/3 x⁶ + ...
= -2x² + ⅔x⁴ - 2x⁴ - 4/45 x⁶ + 4/3 x⁶ - 8/3 x⁶ + ...
= -2x² - 4/3 x⁴ - 64/45 x⁶ + ...

Now, ln x is undefined for x ≤ 0. Likewise, ln(cos2x) is undefined when cos2x ≤ 0
0 ≤ x ≤ ¼π → 0 ≤ 2x ≤ ½π
For this domain, cos2x is always positive, EXCEPT :
When x = ¼π, 2x = ½π
cos2x = cos(½π) = 0

So ln(cos2x) is undefined when x = ¼π, which in turn means that the expansion is not valid.