Obtain Taylor's series expansion for log(cosx) about the point x = π/3 upto the fourth degree term.

To obtain the Taylor series expansion for \( \log(\cos x) \) about \( x = \pi/3 \) up to the fourth-degree term, follow these steps:

### Step 1: General formula for Taylor series expansion
The Taylor series expansion of a function \( f(x) \) about \( x = a \) is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \frac{f^{(4)}(a)}{4!}(x-a)^4 + \cdots \]

Here, \( f(x) = \log(\cos x) \) and \( a = \pi/3 \). We need to compute the derivatives of \( f(x) \) and evaluate them at \( x = \pi/3 \).

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### Step 2: Derivatives of \( f(x) = \log(\cos x) \)

1. **First derivative**:
\[ f'(x) = \frac{d}{dx}[\log(\cos x)] = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \]

2. **Second derivative**:
\[ f''(x) = \frac{d}{dx}[-\tan x] = -\sec^2 x \]

3. **Third derivative**:
\[ f^{(3)}(x) = \frac{d}{dx}[-\sec^2 x] = -2 \sec^2 x \tan x \]

4. **Fourth derivative**:
\[ f^{(4)}(x) = \frac{d}{dx}[-2 \sec^2 x \tan x] = -2 \cdot [2 \sec^2 x \tan^2 x + \sec^4 x] = -2(2 \sec^2 x \tan^2 x + \sec^4 x) \]

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### Step 3: Evaluate derivatives at \( x = \pi/3 \)

At \( x = \pi/3 \), we know:
- \( \cos(\pi/3) = 1/2 \),
- \( \sin(\pi/3) = \sqrt{3}/2 \),
- \( \tan(\pi/3) = \sqrt{3} \),
- \( \sec(\pi/3) = 2 \).

1. \( f(\pi/3) = \log(\cos(\pi/3)) = \log(1/2) = -\log 2 \).

2. \( f'(\pi/3) = -\tan(\pi/3) = -\sqrt{3} \).

3. \( f''(\pi/3) = -\sec^2(\pi/3) = -(2^2) = -4 \).

4. \( f^{(3)}(\pi/3) = -2 \sec^2(\pi/3) \tan(\pi/3) = -2(4)(\sqrt{3}) = -8\sqrt{3} \).

5. \( f^{(4)}(\pi/3) = -2[2 \sec^2(\pi/3) \tan^2(\pi/3) + \sec^4(\pi/3)] = -2[2(4)(3) + 16] = -2[24 + 16] = -80 \).

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### Step 4: Write the Taylor series expansion

Using the Taylor series formula:
\[
f(x) = f(\pi/3) + f'(\pi/3)(x - \pi/3) + \frac{f''(\pi/3)}{2!}(x - \pi/3)^2 + \frac{f^{(3)}(\pi/3)}{3!}(x - \pi/3)^3 + \frac{f^{(4)}(\pi/3)}{4!}(x - \pi/3)^4
\]

Substitute the values:
\[
f(x) = -\log 2 - \sqrt{3}(x - \pi/3) - \frac{4}{2}(x - \pi/3)^2 - \frac{8\sqrt{3}}{6}(x - \pi/3)^3 - \frac{80}{24}(x - \pi/3)^4
\]

Simplify the coefficients:
\[
f(x) = -\log 2 - \sqrt{3}(x - \pi/3) - 2(x - \pi/3)^2 - \frac{4\sqrt{3}}{3}(x - \pi/3)^3 - \frac{10}{3}(x - \pi/3)^4
\]

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### Final Answer:
The Taylor series expansion of \( \log(\cos x) \) about \( x = \pi/3 \) up to the fourth-degree term is:
\[
f(x) = -\log 2 - \sqrt{3}(x - \pi/3) - 2(x - \pi/3)^2 - \frac{4\sqrt{3}}{3}(x - \pi/3)^3 - \frac{10}{3}(x - \pi/3)^4
\]