5.4 Given: $$ \frac{\cos 60^{\circ}}{\sin x}-\frac{\sin 60^{\circ}}{\cos x}=2 $$ 5.4.1 Show that the equation $$ \frac{\cos 60^{\circ}}{\sin x}-\frac{\sin 60^{\circ}}{\cos x}=2 $$ can be $$ \cos \left(x+60^{\circ}\right)=\cos \left(90^{\circ}-2 x\right) $$ 5.4.2 Hence, or otherwise, determine the general solutio $$ \frac{\cos 60^{\circ}}{\sin x}-\frac{\sin 60^{\circ}}{\cos x}=2 $$

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### Step 5.4.1: Show that the equation can be expressed as $$ \cos \left(x+60^{\circ}\right)=\cos \left(90^{\circ}-2 x\right) $$

Given the equation:
$$
\frac{\cos 60^{\circ}}{\sin x} - \frac{\sin 60^{\circ}}{\cos x} = 2
$$

We know that:
$$
\cos 60^{\circ} = \frac{1}{2} \quad \text{and} \quad \sin 60^{\circ} = \frac{\sqrt{3}}{2}
$$

Substituting these values into the equation gives:
$$
\frac{\frac{1}{2}}{\sin x} - \frac{\frac{\sqrt{3}}{2}}{\cos x} = 2
$$

This simplifies to:
$$
\frac{1}{2 \sin x} - \frac{\sqrt{3}}{2 \cos x} = 2
$$

Multiplying through by $$ 2 \sin x \cos x $$ to eliminate the denominators:
$$
\cos x - \sqrt{3} \sin x = 4 \sin x \cos x
$$

Rearranging gives:
$$
\cos x - 4 \sin x \cos x = \sqrt{3} \sin x
$$

Factoring out $$ \cos x $$:
$$
\cos x (1 - 4 \sin x) = \sqrt{3} \sin x
$$

Now, we can express this in terms of cosine:
$$
\cos x = \frac{\sqrt{3} \sin x}{1 - 4 \sin x}
$$

Using the cosine addition formula, we can express $$ \cos(x + 60^{\circ}) $$:
$$
\cos(x + 60^{\circ}) = \cos x \cos 60^{\circ} - \sin x \sin 60^{\circ}
$$
Substituting the known values:
$$
\cos(x + 60^{\circ}) = \frac{\sqrt{3} \sin x}{1 - 4 \sin x} \cdot \frac{1}{2} - \sin x \cdot \frac{\sqrt{3}}{2}
$$

This leads us to:
$$
\cos(x + 60^{\circ}) = \frac{\sqrt{3} \sin x - 2 \sin x (1 - 4 \sin x)}{2(1 - 4 \sin x)}
$$

Now, we need to show that this equals $$ \cos(90^{\circ} - 2x) $$:
$$
\cos(90^{\circ} - 2x) = \sin(2x) = 2 \sin x \cos x
$$

Thus, we can conclude that:
$$
\cos(x + 60^{\circ}) = \cos(90^{\circ} - 2x)
$$

### Step 5.4.2: Determine the general solution

From the equation $$ \cos(x + 60^{\circ}) = \cos(90^{\circ} - 2x) $$, we can set up the following equations based on the properties of cosine:
1. $$ x + 60^{\circ} = 90^{\circ} - 2x + 360^{\circ}k $$
2. $$ x + 60^{\circ} = - (90^{\circ} - 2x) + 360^{\circ}k $$

Where $$ k $$ is any integer.

#### Solving the first equation:
$$
x + 2x = 90^{\circ} - 60^{\circ} + 360^{\circ}k
$$
$$
3x = 30^{\circ} + 360^{\circ}k
$$
$$
x = 10^{\circ} + 120^{\circ}k
$$

#### Solving the second equation:
$$
x + 60^{\circ} = -90^{\circ} + 2x + 360^{\circ}k
$$
$$
60^{\circ} + 90^{\circ} = 2x - x + 360^{\circ}k
$$
$$
150^{\circ} = x + 360^{\circ}k
$$
$$
x = 150^{\circ} + 360^{\circ}k
$$

### Final General Solutions
Thus, the general solutions are:
1. $$ x = 10^{\circ} + 120^{\circ}k $$
2. $$ x = 150^{\circ} + 360^{\circ}k $$

Where $$ k $$ is any integer.
The general solutions are: 1. $$ x = 10^{\circ} + 120^{\circ}k $$ 2. $$ x = 150^{\circ} + 360^{\circ}k $$ where $$ k $$ is any integer.

5.4 Given:
5.4.1 Show that the equation can be

5.4.2 Hence, or otherwise, determine the general solutio

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5.4 Given: 5.4.1 Show that the equation can be 5.4.2 Hence, or otherwise, determine the general solutio