Determine the general solution for the following: 1.1 sin2 x - sin x = cos2 x 1.2 sin2 x + sin x cos x = 0

Solve the quadratic equation $2y^2 - y - 1 = 0$.
Factoring the quadratic:

$$(2y + 1)(y - 1) = 0$$

This gives two possible values for $y$:

$$2y + 1 = 0 \implies y = -\frac{1}{2}$$ $$y - 1 = 0 \implies y = 1$$

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

For $\sin x = 1$:
The principal value is $x = 90^\circ$.
The general solution is $x = 90^\circ + 360^\circ k$, where $k \in \mathbb{Z}$.

For $\sin x = -\frac{1}{2}$:
The reference angle is $30^\circ$. Since $\sin x$ is negative, $x$ lies in the third or fourth quadrants.
In the third quadrant: $x = 180^\circ + 30^\circ = 210^\circ$.
In the fourth quadrant: $x = 360^\circ - 30^\circ = 330^\circ$.
The general solutions are $x = 210^\circ + 360^\circ k$ and $x = 330^\circ + 360^\circ k$, where $k \in \mathbb{Z}$.

The general solution is $\boxed{x = 90^\circ + 360^\circ k, x = 210^\circ + 360^\circ k, x = 330^\circ + 360^\circ k, k \in \mathbb{Z}}$.

Case 1: $\sin x = 0$
The general solution is $x = 180^\circ k$, where $k \in \mathbb{Z}$.

Case 2: $\sin x + \cos x = 0$
Divide by $\cos x$ (assuming $\cos x \neq 0$. If $\cos x = 0$, then $\sin x = \pm 1$, so $\sin x + \cos x = \pm 1 \neq 0$. Thus $\cos x \neq 0$ is a valid assumption).

$$\frac{\sin x}{\cos x} + \frac{\cos x}{\cos x} = 0$$ $$\tan x + 1 = 0$$ $$\tan x = -1$$

The principal value for $\tan x = -1$ is $x = -45^\circ$ or $x = 135^\circ$.
The general solution is $x = 135^\circ + 180^\circ k$, where $k \in \mathbb{Z}$.

The general solution is $\boxed{x = 180^\circ k, x = 135^\circ + 180^\circ k, k \in \mathbb{Z}}$.

2. Determine the general solution of $3\cos 2x = 1 + 5\cos x$

$$6\cos^2 x - 3 = 1 + 5\cos x$$ $$6\cos^2 x - 5\cos x - 4 = 0$$

Solve the quadratic equation $6y^2 - 5y - 4 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(6)(-4)}}{2(6)}$$ $$y = \frac{5 \pm \sqrt{25 + 96}}{12}$$ $$y = \frac{5 \pm \sqrt{121}}{12}$$ $$y = \frac{5 \pm 11}{12}$$

This gives two possible values for $y$:

$$y_1 = \frac{5 + 11}{12} = \frac{16}{12} = \frac{4}{3}$$ $$y_2 = \frac{5 - 11}{12} = \frac{-6}{12} = -\frac{1}{2}$$

So, $\cos x = \frac{4}{3}$ or $\cos x = -\frac{1}{2}$.

For $\cos x = \frac{4}{3}$: This has no solution because the range of $\cos x$ is $[-1, 1]$.
For $\cos x = -\frac{1}{2}$:
The principal value is $x = 120^\circ$.
The general solution is $x = \pm 120^\circ + 360^\circ k$, where $k \in \mathbb{Z}$.

The general solution is $\boxed{x = \pm 120^\circ + 360^\circ k, k \in \mathbb{Z}}$.

3. Solve for $x$: $4\sin^2 x - 3\sin x - 1 = 0$ for $x \in [0^\circ; 360^\circ]$

Let $y = \sin x$.

$$4y^2 - 3y - 1 = 0$$

Factoring the quadratic:

$$(4y + 1)(y - 1) = 0$$

This gives two possible values for $y$:

$$4y + 1 = 0 \implies y = -\frac{1}{4}$$ $$y - 1 = 0 \implies y = 1$$

So, $\sin x = -\frac{1}{4}$ or $\sin x = 1$.

For $\sin x = 1$:

$$x = 90^\circ$$

For $\sin x = -\frac{1}{4}$:
The reference angle is $\alpha = \arcsin\left(\frac{1}{4}\right) \approx 14.48^\circ$.
Since $\sin x$ is negative, $x$ lies in the third or fourth quadrants.
In the third quadrant: $x = 180^\circ + \alpha \approx 180^\circ + 14.48^\circ = 194.48^\circ$.
In the fourth quadrant: $x = 360^\circ - \alpha \approx 360^\circ - 14.48^\circ = 345.52^\circ$.

The values of $x$ in the given range are $\boxed{x = 90^\circ, 194.48^\circ, 345.52^\circ}$.

4. Determine the general solution of $\tan x = 2\sin 2x$ where $\cos x < 0$.

$$\frac{\sin x}{\cos x} = 2(2\sin x \cos x)$$ $$\frac{\sin x}{\cos x} = 4\sin x \cos x$$

$$\sin x = 4\sin x \cos^2 x$$ $$\sin x - 4\sin x \cos^2 x = 0$$ $$\sin x (1 - 4\cos^2 x) = 0$$

Case 1: $\sin x = 0$
The general solution is $x = 180^\circ k$, where $k \in \mathbb{Z}$.
We also need to consider the condition $\cos x < 0$.
If $x = 180^\circ k$:
If $k$ is even, $x = 0^\circ, 360^\circ, \dots$, then $\cos x = 1$. This does not satisfy $\cos x < 0$.
If $k$ is odd, $x = 180^\circ, 540^\circ, \dots$, then $\cos x = -1$. This satisfies $\cos x < 0$.
So, for this case, $x = 180^\circ (2m+1)$, where $m \in \mathbb{Z}$. This can be written as $x = 180^\circ + 360^\circ m$.

Case 2: $1 - 4\cos^2 x = 0$

$$4\cos^2 x = 1$$ $$\cos^2 x = \frac{1}{4}$$ $$\cos x = \pm \frac{1}{2}$$

We are given the condition $\cos x < 0$, so we must have $\cos x = -\frac{1}{2}$.
For $\cos x = -\frac{1}{2}$:
The principal value is $x = 120^\circ$.
The general solution is $x = \pm 120^\circ + 360^\circ k$, where $k \in \mathbb{Z}$.
Both $120^\circ$ and $-120^\circ$ (or $240^\circ$) have negative cosine values.
So, $x = 120^\circ + 360^\circ k$ and $x = 240^\circ + 360^\circ k$ (or $x = -120^\circ + 360^\circ k$).

Combining the valid solutions from both cases:
The general solution is $\boxed{x = 180^\circ + 360^\circ k, x = 120^\circ + 360^\circ k, x = 240^\circ + 360^\circ k, k \in \mathbb{Z}}$.

5. Calculate the value of $x$, if $x \in [-180^\circ; 360^\circ]$ for $\cos 2x = \cos x + 2$

Solve the quadratic equation $2y^2 - y - 3 = 0$.
Factoring the quadratic:

$$(2y - 3)(y + 1) = 0$$

This gives two possible values for $y$:

$$2y - 3 = 0 \implies y = \frac{3}{2}$$ $$y + 1 = 0 \implies y = -1$$

So, $\cos x = \frac{3}{2}$ or $\cos x = -1$.

For $\cos x = \frac{3}{2}$: This has no solution because the range of $\cos x$ is $[-1, 1]$.
For $\cos x = -1$:
The values of $x$ in the range $[-180^\circ; 360^\circ]$ are:

$$x = 180^\circ$$ $$x = -180^\circ$$

The values of $x$ in the given range are $\boxed{x = -180^\circ, 180^\circ}$.