$$ \sin x \quad \cos x $$ 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 solution $$ \frac{\cos 60^{\circ}}{\sin x}-\frac{\sin 60^{\circ}}{\cos x}=2 $$

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We start with
$$
\frac{\cos 60^\circ}{\sin x} - \frac{\sin 60^\circ}{\cos x} = 2.
$$

Since
$$
\cos 60^\circ = \frac{1}{2} \quad \text{and} \quad \sin 60^\circ = \frac{\sqrt{3}}{2},
$$
the equation becomes
$$
\frac{\frac{1}{2}}{\sin x} - \frac{\frac{\sqrt{3}}{2}}{\cos x} = 2.
$$

Multiply both sides by 2:
$$
\frac{1}{\sin x} - \frac{\sqrt{3}}{\cos x} = 4.
$$

Next, multiply through by $$\sin x \cos x$$ (assuming $$\sin x \neq 0$$ and $$\cos x \neq 0$$):
$$
\cos x - \sqrt{3}\sin x = 4\sin x \cos x.
$$

Observe that
$$
4\sin x \cos x = 2\sin 2x,
$$
since $$\sin 2x = 2\sin x \cos x$$. Thus, we have
$$
\cos x - \sqrt{3} \sin x = 2\sin 2x.
$$

Now, notice that using the cosine addition formula,
$$
\cos (x+60^\circ) = \cos x \cos 60^\circ - \sin x \sin 60^\circ = \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x.
$$
Multiplying by 2 gives
$$
2\cos (x+60^\circ) = \cos x - \sqrt{3}\sin x.
$$

Replacing the left-hand side in our equation leads to
$$
2\cos(x+60^\circ) = 2\sin 2x.
$$
Dividing both sides by 2 yields
$$
\cos(x+60^\circ) = \sin 2x.
$$

Since
$$
\sin 2x = \cos(90^\circ-2x),
$$
we obtain
$$
\cos(x+60^\circ) = \cos(90^\circ-2x).
$$

This completes part 5.4.1.

---

For 5.4.2, we solve
$$
\cos(x+60^\circ) = \cos(90^\circ-2x).
$$

Recall that if $$\cos A = \cos B$$ then
$$
A = 360^\circ k \pm B, \quad k\in\mathbb{Z}.
$$

We consider two cases.

**Case 1:**
$$
x + 60^\circ = 90^\circ - 2x + 360^\circ k.
$$
Solve for $$x$$:
$$
x + 2x = 90^\circ - 60^\circ + 360^\circ k,
$$
$$
3x = 30^\circ + 360^\circ k,
$$
$$
x = 10^\circ + 120^\circ k.
$$

**Case 2:**
$$
x + 60^\circ = -\bigl(90^\circ - 2x\bigr) + 360^\circ k.
$$
Simplify:
$$
x + 60^\circ = -90^\circ + 2x + 360^\circ k.
$$
Rearrange:
$$
60^\circ + 90^\circ = 2x - x + 360^\circ k,
$$
$$
150^\circ = x + 360^\circ k,
$$
$$
x = 150^\circ - 360^\circ k.
$$
Since $$k$$ takes all integer values, we can write this as
$$
x = 150^\circ + 360^\circ k, \quad k\in\mathbb{Z}.
$$

Thus, the general solution is:
$$
x = 10^\circ + 120^\circ k \quad \text{or} \quad x = 150^\circ + 360^\circ k, \quad k\in\mathbb{Z}.
$$
The general solution to the equation $$ \frac{\cos 60^\circ}{\sin x} - \frac{\sin 60^\circ}{\cos x} = 2 $$ is: $$ x = 10^\circ + 120^\circ k \quad \text{or} \quad x = 150^\circ + 360^\circ k, \quad \text{where } k \text{ is any integer.} $$

5.4.1 Show that the equation can be

5.4.2 Hence, or otherwise, determine the general solution

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5.4.1 Show that the equation can be 5.4.2 Hence, or otherwise, determine the general solution