(A)(i) Start with:

$$g(x)=2\ln(x^7)-\left(3\ln x+\ln(8x^4)\right).$$

Use the power rule:

$$2\ln(x^7)=\ln\left((x^7)^2\right)=\ln(x^{14}).$$

Rewrite the subtraction:

$$g(x)=\ln(x^{14})-3\ln x-\ln(8x^4).$$

Use the power rule again:

$$3\ln x=\ln(x^3).$$

Now:

$$g(x)=\ln(x^{14})-\ln(x^3)-\ln(8x^4).$$

Combine with the quotient rule:

$$g(x)=\ln\left(\frac{x^{14}}{x^3(8x^4)}\right).$$

Simplify:

$$g(x)=\ln\left(\frac{x^{14}}{8x^7}\right)=\ln\left(\frac{x^7}{8}\right).$$

Final answer: $$\ln\left(\frac{x^7}{8}\right)$$

(A)(ii) Start with:

$$h(x)=\frac{\sin 2x}{1-\cos^2 x}.$$

Use the identity:

$$1-\cos^2 x=\sin^2 x.$$

Then:

$$h(x)=\frac{\sin 2x}{\sin^2 x}.$$

Use the double-angle identity:

$$\sin 2x=2\sin x\cos x.$$

Substitute:

$$h(x)=\frac{2\sin x\cos x}{\sin^2 x}.$$

Simplify:

$$h(x)=2\frac{\cos x}{\sin x}=2\cot x.$$

Final answer: $$2\cot x$$

(B)(i) Solve:

$$\tan^2 x-1=0.$$

Then:

$$\tan^2 x=1,$$

$$\tan x=\pm1.$$

On the interval $$[0,\pi]$$:

$$\tan x=1$$ at $$x=\frac{\pi}{4},$$

$$\tan x=-1$$ at $$x=\frac{3\pi}{4}.$$

Final answers: $$x=\frac{\pi}{4},\frac{3\pi}{4}$$

(B)(ii) Interpret:

$$\frac{1}{e^{-3}}=e^3.$$

So solve:

$$5e^{x-2}=e^3.$$

Divide by $$5$$:

$$e^{x-2}=\frac{e^3}{5}.$$

Take natural logarithms:

$$x-2=\ln\left(\frac{e^3}{5}\right)=3-\ln 5.$$

Solve for $$x$$:

$$x=5-\ln 5.$$

Compare with:

$$x=a-\ln a.$$

So:

$$a=5.$$

Final answer: $$a=5$$

$$\cos 2x+3\sin x=-1.$$

Use:

$$\cos 2x=1-2\sin^2 x.$$

Substitute:

$$1-2\sin^2 x+3\sin x=-1.$$

Move all terms to one side:

$$-2\sin^2 x+3\sin x+2=0.$$

Multiply by $$-1$$:

$$2\sin^2 x-3\sin x-2=0.$$

Let:

$$u=\sin x.$$

Then:

$$2u^2-3u-2=0.$$

Factor:

$$(2u+1)(u-2)=0.$$

So:

$$u=-\frac{1}{2}$$

$$u=2.$$

Since sine cannot equal $$2$$, keep:

$$\sin x=-\frac{1}{2}.$$

The solutions are:

$$x=\frac{7\pi}{6}+2\pi k,\frac{11\pi}{6}+2\pi k,$$ where $$k$$ is any integer.

Final answers:

$$x=\frac{7\pi}{6}+2\pi k,\frac{11\pi}{6}+2\pi k,\quad k\in\mathbb{Z}$$

Section II Part B

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