5.1 simplify the following expression to a single trigonometric ratio: \( \frac{\sin \left(180^{\circ}-\beta\right) \cdot \sin \beta-1}{\cos (-\beta)} \) 5.2 Hence, determine for which value(s) of \( \beta \), where \( \beta \in\left[0^{\circ} ; 360^{\circ}\right] \) \( \frac{\sin \left(180^{\circ}-\beta\right) \cdot \sin \beta-1}{\cos (-\beta)} \) will be undefined. 5.3 Determine the general solution of \( 3 \sin x=2 \tan x \)

Let's address each part of the problem step-by-step. --- ### **5.1 Simplify the Expression** **Given Expression:** \[ \frac{\sin \left(180^{\circ}-\beta\right) \cdot \sin \beta - 1}{\cos (-\beta)} \] **Simplification Steps:** 1. **Apply Trigonometric Identities:** - \(\sin(180^{\circ} - \beta) = \sin \beta\) (since \(\sin\) is positive in the second quadrant). - \(\cos(-\beta) = \cos \beta\) (since \(\cos\) is an even function). 2. **Substitute the Identities:** \[ \frac{\sin \beta \cdot \sin \beta - 1}{\cos \beta} = \frac{\sin^2 \beta - 1}{\cos \beta} \] 3. **Use the Pythagorean Identity:** \[ \sin^2 \beta + \cos^2 \beta = 1 \quad \Rightarrow \quad \sin^2 \beta - 1 = -\cos^2 \beta \] 4. **Substitute and Simplify:** \[ \frac{-\cos^2 \beta}{\cos \beta} = -\cos \beta \] **Final Simplified Form:** \[ -\cos \beta \] --- ### **5.2 Determine When the Expression is Undefined** **Expression After Simplification:** \[ -\cos \beta \] **Original Expression:** \[ \frac{\sin^2 \beta - 1}{\cos \beta} \] **Understanding When It's Undefined:** The expression is undefined when the denominator is zero. Therefore, we need to find the values of \(\beta\) where: \[ \cos \beta = 0 \] **Solutions in the Interval \([0^{\circ}, 360^{\circ}]\):** \[ \cos \beta = 0 \quad \Rightarrow \quad \beta = 90^{\circ}, \ 270^{\circ} \] **Final Answer:** The expression is undefined at \(\beta = 90^{\circ}\) and \(\beta = 270^{\circ}\). --- ### **5.3 Determine the General Solution of \(3 \sin x = 2 \tan x\)** **Given Equation:** \[ 3 \sin x = 2 \tan x \] **Step-by-Step Solution:** 1. **Express \(\tan x\) in Terms of \(\sin x\) and \(\cos x\):** \[ \tan x = \frac{\sin x}{\cos x} \] Substituting: \[ 3 \sin x = 2 \left( \frac{\sin x}{\cos x} \right) \] 2. **Assume \(\sin x \neq 0\) and \(\cos x \neq 0\) (we'll address \(\sin x = 0\) later):** \[ 3 \sin x \cos x = 2 \sin x \] Divide both sides by \(\sin x\): \[ 3 \cos x = 2 \] Solve for \(\cos x\): \[ \cos x = \frac{2}{3} \] 3. **Find Solutions When \(\cos x = \frac{2}{3}\):** \[ x = \pm \cos^{-1}\left( \frac{2}{3} \right) + 360^{\circ}k, \quad k \in \mathbb{Z} \] 4. **Consider the Case When \(\sin x = 0\):** Substitute \(\sin x = 0\) into the original equation: \[ 3(0) = 2 \tan x \quad \Rightarrow \quad 0 = 0 \] This is always true. Therefore, all angles where \(\sin x = 0\) are solutions: \[ x = 180^{\circ}k, \quad k \in \mathbb{Z} \] **Combined General Solution:** \[ x = 180^{\circ}k \quad \text{or} \quad x = \pm \cos^{-1}\left( \frac{2}{3} \right) + 360^{\circ}k, \quad k \in \mathbb{Z} \] **Final Answer:** All real numbers congruent to 0° modulo 180°, and all real numbers equal to cos⁻¹ (2⁄3) plus a multiple of 360°. In other words, the general solutions are  x = 180°·k  or  x = cos⁻¹(2/3) + 360°·k  for any integer k.