Let \(u = \cos(3x-4)\), so that
$$
u + 1 = -u
$$
or
$$
u = \frac{-1}{2}.
$$
So,
$$
\cos (3x-4) = \frac{-1}{2}
$$
From the unit circle, we have that
$$
3x-4 = \pm\frac{5\pi}{6}+2n\pi,
$$
or
$$
x = \pm\frac{5\pi}{18}+\frac{4}{3}+\frac{2}{3}n\pi.
$$
[qed]

Let \(u = \tan(4x-\pi)\). Then
$$
\frac{7}{4}u^2-\frac{5}{2}= -\frac{9}{4}
$$
or
$$
u = \pm \frac{1}{\sqrt{7}}
$$
So,
$$
\tan(4x-\pi) = \pm \frac{1}{\sqrt{7}}.
$$

The unit circle won’t help here, so we make use of inverse trigonometric functions. So,
$$
4x-\pi = \tan^{-1}(1/\sqrt{7})
$$
or
$$
4x-\pi = \tan^{-1}(-1/\sqrt{7})
$$
That is,
$$
x = \frac{\tan^{-1}(1/\sqrt{7})+\pi}{4} + \frac{1}{4}n\pi
$$
or
$$
x = \frac{\tan^{-1}(-1/\sqrt{7})+\pi}{4} + \frac{1}{4}n\pi
$$
[qed]

Let \(u = \cot x\) and \(v = \cos x\), then
$$
u v^2 – \frac{1}{2}u = 0
$$
Now factor.
$$
u \left(v^2- \frac{1}{2} \right) = 0
$$
That is, either
$$
u = 0 \text{ or } v^2 -\frac{1}{2} = 0.
$$
That is, either
$$
\cot x = 0 \text{ or } \cos x = \pm \frac{1}{\sqrt{2}}.
$$
That is, either
$$
x = \pm\frac{\pi}{2} + 2 n \pi
$$
or
$$
x = \pm \frac{\pi}{4} + 2n \pi
$$
or
$$
x = \pm \frac{3\pi}{4} + 2n \pi
$$
[qed]

Let \(u = \sin x\). Then
$$
2u^2 – u – 1 = 0.
$$
So,
$$
(2u+1)(u-1) = 0
$$
Therefore, either
$$
u = \frac{-1}{2} \text{ or } u = 1.
$$

So, either
$$
\sin x = \frac{-1}{2} \text{ or } \sin x = 1.
$$

That is, either
$$
x = \frac{-\pi}{6} + 2n\pi
$$
or
$$
x = \frac{-5\pi}{6} + 2n \pi
$$
or
$$
x = \frac{\pi}{2} + 2n\pi
$$
[qed]

Recall that \(\sin^2x + \cos^2 x = 1\), or \(\cos^2 x = 1-\sin^2x\).
So the given equation can be expressed as
$$
2(1-\sin^2 x) + 3 \sin x -3 =0
$$
or
$$
2\sin^2x -3 \sin x +1=0.
$$
Let \(u = \sin x.\) Then
$$
2u^2 – 3u +1 = 0
$$
or
$$
(2u-1)(u-1) = 0.
$$
That is,
$$
u = \frac{1}{2} \text{ or } u = 1.
$$
That is,
$$
\sin x = \frac{1}{2} \text{ or } \sin x = 1.
$$
That is, either
$$
x = \frac{\pi}{6} + 2n\pi
$$
or
$$
x = \frac{5\pi}{6} + 2n \pi
$$
or
$$
x = \frac{3\pi}{2} + 2n\pi
$$
[qed]

[qed]