1. 2sin^2 x - 5sin x + 2 = 0 2. cos^2 x + 3cos x + 2 = 0 3. 2sin^2 x - 3sin x + 1 = 0 ; 4. tg2x-3tgx+2=0; 5. 2cos^2 x - 3cos x - 2 = 0 6. 2cos^2 x + sqrt(2) * cos x = 0 7.…

To solve these trigonometric equations, we'll apply algebraic techniques and relationships between trigonometric functions. Below are the solutions for the given equations:

1. \( 2\sin^2 x - 5\sin x + 2 = 0 \)

Let \( y = \sin x \): \[ 2y^2 - 5y + 2 = 0 \] Using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \] \[ y = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \] Thus, \( y = 2 \) or \( y = \frac{1}{2} \). However, \( \sin x \) cannot be greater than 1, so we discard \( y = 2 \).

For \( \sin x = \frac{1}{2} \): \[ x = \frac{\pi}{6} + 2k\pi \text{ or } x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z} \]

2. \( \cos^2 x + 3\cos x + 2 = 0 \)

Let \( z = \cos x \): \[ z^2 + 3z + 2 = 0 \] Factoring: \[ (z + 1)(z + 2) = 0 \] Thus, \( z = -1 \) or \( z = -2 \). \( \cos x = -2 \) is not valid.

For \( \cos x = -1 \): \[ x = \pi + 2k\pi, \quad k \in \mathbb{Z} \]

3. \( 2\sin^2 x - 3\sin x + 1 = 0 \)

Let \( y = \sin x \): \[ 2y^2 - 3y + 1 = 0 \] Using the quadratic formula: \[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm 1}{4} \] So, \( y = 1 \text{ or } y = \frac{1}{2} \).

For \( \sin x = 1 \): \[ x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z} \]

For \( \sin x = \frac{1}{2} \): \[ x = \frac{\pi}{6} + 2k\pi \text{ or } x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z} \]

4. \( \tan 2x - 3\tan x + 2 = 0 \)

Let \( z = \tan x \): Using the double angle formula \( \tan 2x = \frac{2z}{1-z^2} \): \[ \frac{2z}{1-z^2} - 3z + 2 = 0 \] Multiplying through by \( 1 - z^2 \): \[ 2z - 3z(1 - z^2) + 2(1 - z^2) = 0 \] Simplifying: \[ 2z - 3z + 3z^3 + 2 - 2z^2 = 0 \Rightarrow 3z^3 - 2z^2 - z + 2 = 0 \] This cubic can be solved numerically or using synthetic division for rational roots.

5. \( 2\cos^2 x - 3\cos x - 2 = 0 \)

Let \( z = \cos x \): \[ 2z^2 - 3z - 2 = 0 \] Using the quadratic formula: \[ z = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 16}}{4} \] \[ z = \frac{3 \pm 5}{4} \] Thus, \( z = 2 \) or \( z = -\frac{1}{2} \).

For \( \cos x = -\frac{1}{2} \): \[ x = \frac{2\pi}{3} + 2k\pi \text{ or } x = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z} \]

6. \( 2\cos^2 x + \sqrt{2}\cos x = 0 \)

Factoring: \[ \cos x(2\cos x + \sqrt{2}) = 0 \] Thus, \( \cos x = 0 \) or \( 2\cos x + \sqrt{2} = 0 \), implying \( \cos x = -\frac{\sqrt{2}}{2} \).

For \( \cos x = 0 \): \[ x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \]

For \( \cos x = -\frac{\sqrt{2}}{2} \): \[ x = \frac{3\pi}{4} + 2k\pi \text{ or } x = \frac{5\pi}{4} + 2k\pi, \quad k \in \mathbb{Z} \]

7. \( \tan x + 4\cot x - 5 = 0 \)

Substituting \( \cot x = \frac{1}{\tan x} \): Let \( z = \tan x \): \[ z + \frac{4}{z} - 5 = 0 \] Multiplying through by \( z \): \[ z^2 - 5z + 4 = 0 \] Using the quadratic formula: \[ z = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2} \] So \( z = 4 \text{ or } z = 1 \).

For \( \tan x = 4 \): \[ x = \tan^{-1}(4) + k\pi, \quad k \in \mathbb{Z} \]

For \( \tan x = 1 \): \[ x = \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z} \]

8. \( \cos^2 x + \sin x + 1 = 0 \)

Substituting \( \sin^2 x = 1 - \cos^2 x \): \[ \cos^2 x + (1 - \cos^2 x) + 1 = 0 \Rightarrow 1 = 0 \quad \text{(No solution)} \]

9. \( 2\cos^2 x + 3\sin x - 3 = 0 \)

Substituting \( \sin^2 x = 1 - \cos^2 x \): \[ 2(1 - \sin^2 x) + 3\sin x - 3 = 0 \] Thus: \[ -2\sin^2 x + 3\sin x - 1 = 0 \] Let \( y = \sin x \): \[ 2y^2 - 3y + 1 = 0 \] Using the quadratic formula: \[ y = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \] So \( y = 1 \text{ or } y = \frac{1}{2} \).

For \( \sin x = 1 \): \[ x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z} \]

For \( \sin x = \frac{1}{2} \): \[ x = \frac{\pi}{6} + 2k\pi \text{ or } \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z} \]

10. \( \tan x - 2\cot x + 1 = 0 \)

Let \( z = \tan x \): \[ z - \frac{2}{z} + 1 = 0 \] Multiplying by \( z \): \[ z^2 + z - 2 = 0 \] Factoring: \[ (z - 1)(z + 2) = 0 \] Thus, \( \tan x = 1 \) or \( \tan x = -2 \).

For \( \tan x = 1 \): \[ x = \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z} \]

For \( \tan x = -2 \): \[ x = \tan^{-1}(-2) + k\pi, \quad k \in \mathbb{Z} \]

Feel free to ask for further clarifications or any specific problems!