Solve the equations ( \( 95= \) gene 1) \( 3 \sin x-2 \cos x=0 \) 2) \( 3 \sin ^{2} x-2 \sin x=0 \) and 3) \( 3 \tan ^{2} x+\tan x=2 \) and -1 4) \( 5 \sin ^{2} x-3 \sin x-2=0 \)

1) We start with \[ 3\sin x - 2\cos x = 0. \] Rearrange the equation: \[ 3\sin x = 2\cos x. \] Assuming \(\cos x \neq 0\), divide both sides by \(\cos x\): \[ 3\tan x = 2 \quad\Longrightarrow\quad \tan x = \frac{2}{3}. \] Thus the general solution is \[ x = \arctan\left(\frac{2}{3}\right) + \pi k,\quad k\in\mathbb{Z}. \] 2) The equation is \[ 3\sin^2 x - 2\sin x = 0. \] Factor out \(\sin x\): \[ \sin x\,(3\sin x - 2) = 0. \] This gives two cases: • Case 1: \[ \sin x = 0 \quad\Longrightarrow\quad x = \pi k,\quad k\in\mathbb{Z}. \] • Case 2: \[ 3\sin x - 2 = 0\quad\Longrightarrow\quad \sin x = \frac{2}{3}. \] The general solution for \(\sin x = \frac{2}{3}\) is \[ x = \arcsin\left(\frac{2}{3}\right) + 2\pi k \quad \text{or} \quad x = \pi - \arcsin\left(\frac{2}{3}\right) + 2\pi k,\quad k\in\mathbb{Z}. \] 3) The equation is \[ 3\tan^2 x + \tan x - 2 = 0. \] Let \(y = \tan x\). Then the quadratic becomes \[ 3y^2 + y - 2 = 0. \] Compute the discriminant: \[ \Delta = 1^2 - 4(3)(-2) = 1 + 24 = 25. \] Solve for \(y\): \[ y = \frac{-1\pm \sqrt{25}}{6} = \frac{-1\pm 5}{6}. \] This gives the solutions: \[ y = \frac{4}{6} = \frac{2}{3} \quad \text{or} \quad y = \frac{-6}{6} = -1. \] Thus we have two cases: • Case 1: \[ \tan x = \frac{2}{3}\quad\Longrightarrow\quad x = \arctan\left(\frac{2}{3}\right) + \pi k,\quad k\in\mathbb{Z}. \] • Case 2: \[ \tan x = -1\quad\Longrightarrow\quad x = -\frac{\pi}{4} + \pi k,\quad k\in\mathbb{Z}. \] 4) The equation is \[ 5\sin^2 x - 3\sin x - 2=0. \] Let \(y = \sin x\). Then the quadratic is \[ 5y^2 - 3y - 2 = 0. \] Compute the discriminant: \[ \Delta = (-3)^2 - 4(5)(-2) = 9 + 40 = 49. \] Solve for \(y\): \[ y = \frac{3 \pm 7}{10}. \] This gives: \[ y = \frac{10}{10} = 1 \quad \text{or} \quad y = \frac{-4}{10} = -0.4. \] Now solve for \(x\): • Case 1: \(\sin x = 1\) gives \[ x = \frac{\pi}{2} + 2\pi k,\quad k\in\mathbb{Z}. \] • Case 2: \(\sin x = -0.4\) gives \[ x = \arcsin(-0.4) + 2\pi k \quad \text{or} \quad x = \pi - \arcsin(-0.4) + 2\pi k,\quad k\in\mathbb{Z}. \] Since \(\arcsin(-0.4) = -\arcsin(0.4)\), the solutions can also be written as \[ x = -\arcsin(0.4) + 2\pi k \quad \text{or} \quad x = \pi + \arcsin(0.4) + 2\pi k,\quad k\in\mathbb{Z}. \]