写出 $\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)$ 的主值， $\tan^{-1}x$ 的主值必须介于其间的数值， 使方程 $4\sin^{2}x = k$ 有解的 $k$ 的取值范围。 证明恒等式 $\cot x - \cot x\tan^{2}x + \tan x = \dfrac{1 + \cos 2x}{\sin 2x}$。 由此解方程 $\cot x - \cot x\tan^{2}x + \tan x = \csc 2x - 3$，其中 $0^\circ \le x \le 180^\circ$。

1. Determine the reference angle

Let $\theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
The reference angle is $\frac{\pi}{3}$ (or $60^\circ$).

2. Apply the principal value range for inverse sine

The principal value range for $\sin^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Since the argument is negative, $\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$ (or $-60^\circ$).

1. State the range of the principal value of $\tan^{-1}x$

The principal value of $\tan^{-1}x$ lies strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ radians (or $-90^\circ$ and $90^\circ$).

1. Determine the range of values for $\sin^2 x$

Since $-1 \le \sin x \le 1$, squaring it gives:
$0 \le \sin^2 x \le 1$

2. Find the range of $k$

Multiply the inequality by 4:
$0 \le 4\sin^2 x \le 4$
Since $4\sin^2 x = k$, we have:
$0 \le k \le 4$

1. Simplify the second term of the left-hand side (LHS).

$\text{LHS} = \cot x - \cot x\tan^{2}x + \tan x$
Since $\cot x = \frac{1}{\tan x}$, $\cot x\tan^{2}x = \tan x$.
$\text{LHS} = \cot x - \tan x + \tan x$

2. Simplify the LHS expression.

$\text{LHS} = \cot x$

3. Express the simplified LHS in terms of sine and cosine, and introduce a factor to match the RHS denominator.

$\text{LHS} = \frac{\cos x}{\sin x}$
Multiply numerator and denominator by $2\cos x$:
$\text{LHS} = \frac{2\cos^{2}x}{2\sin x\cos x}$

4. Apply double angle formulae to obtain the RHS.

Using $\cos 2x = 2\cos^{2}x - 1 \implies 2\cos^{2}x = 1 + \cos 2x$ and $\sin 2x = 2\sin x\cos x$:
$\text{LHS} = \frac{1 + \cos 2x}{\sin 2x} = \text{RHS}$