6. Solve: a) \( \sin \theta \cdot \cos 30^{\circ}+\cos \theta \cdot \sin 30^{\circ}=0,5 \) b) \( \sin \theta \cdot \sin 20^{\circ}+\cos \theta \cdot \cos 20^{\circ}=\sin 50^{\circ} \) Find \( \theta \in\left[-180^{\circ} ; 180^{\circ}\right] \operatorname{such} \) that: a) \( \sin 5 \theta \cdot \cos 20^{\circ}-\cos 5 \theta \cdot \sin 20^{\circ}=1 \) b) \( 2 \cos 3 \theta \cdot \cos 30^{\circ}-2 \sin 3 \theta \cdot \sin 30^{\circ}=1 \)

We first note that both parts consist of rewriting the given expressions by using trigonometric addition formulas. --- ### 1(a) We are to solve \[ \sin \theta \cdot \cos 30^{\circ} + \cos \theta \cdot \sin 30^{\circ} = 0.5. \] Using the sine addition formula \[ \sin(\theta + 30^{\circ}) = \sin \theta \cos 30^{\circ} + \cos \theta \sin 30^{\circ}, \] the equation becomes \[ \sin (\theta + 30^{\circ}) = 0.5. \] Since \(\sin \alpha = 0.5\) when \[ \alpha = 30^{\circ} + 360^{\circ} k\quad \text{or}\quad \alpha = 150^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z}, \] we set \[ \theta + 30^{\circ} = 30^{\circ} + 360^{\circ} k\quad \text{or}\quad \theta + 30^{\circ} = 150^{\circ} + 360^{\circ} k. \] Solving these: 1. For \(\theta + 30^{\circ} = 30^{\circ} + 360^{\circ} k\): \[ \theta = 360^{\circ} k. \] 2. For \(\theta + 30^{\circ} = 150^{\circ} + 360^{\circ} k\): \[ \theta = 120^{\circ} + 360^{\circ} k. \] Thus, the solutions are \[ \theta = 360^{\circ} k \quad \text{or} \quad \theta = 120^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z}. \] --- ### 1(b) We now solve \[ \sin \theta \cdot \sin 20^{\circ} + \cos \theta \cdot \cos 20^{\circ}=\sin 50^{\circ}. \] Note that the cosine addition formula is \[ \cos(\theta - 20^{\circ}) = \cos \theta \cos 20^{\circ} + \sin \theta \sin 20^{\circ}. \] Thus, the left-hand side becomes \[ \cos(\theta - 20^{\circ}). \] So the equation is \[ \cos(\theta - 20^{\circ}) = \sin 50^{\circ}. \] Recognize that \(\sin50^{\circ}=\cos40^{\circ}\). Therefore, the equation is \[ \cos(\theta - 20^{\circ}) = \cos 40^{\circ}. \] For \(\cos \alpha = \cos \beta\), the general solution is \[ \alpha = \beta + 360^{\circ} k \quad \text{or} \quad \alpha = -\beta + 360^{\circ} k,\quad k \in\mathbb{Z}. \] Thus, setting \(\alpha = \theta-20^{\circ}\) and \(\beta = 40^{\circ}\): 1. \(\theta - 20^{\circ} = 40^{\circ} + 360^{\circ} k\) gives \[ \theta = 60^{\circ} + 360^{\circ} k. \] 2. \(\theta - 20^{\circ} = -40^{\circ} + 360^{\circ} k\) gives \[ \theta = -20^{\circ} + 360^{\circ} k. \] Thus, the solutions are \[ \theta = 60^{\circ} + 360^{\circ} k\quad \text{or}\quad \theta = -20^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z}. \] --- Now we turn to the next two problems where we also require that \(\theta \in [-180^{\circ}, 180^{\circ}]\). --- ### 2(a) We wish to find \(\theta \in [-180^{\circ}, 180^{\circ}]\) such that \[ \sin 5\theta \cdot \cos 20^{\circ} - \cos 5\theta \cdot \sin 20^{\circ} = 1. \] Recognize that \[ \sin 5\theta \cos 20^{\circ} - \cos 5\theta \sin 20^{\circ} = \sin(5\theta - 20^{\circ}) \] by the sine subtraction formula. Thus, the equation becomes \[ \sin(5\theta - 20^{\circ}) = 1. \] We know that \(\sin \alpha = 1\) when \[ \alpha = 90^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z}. \] So set \[ 5\theta - 20^{\circ}= 90^{\circ} + 360^{\circ} k. \] Solving for \(\theta\): \[ 5\theta = 110^{\circ} + 360^{\circ} k, \] \[ \theta = 22^{\circ} + 72^{\circ} k. \] We must choose integers \(k\) so that \(\theta\) lies in \([-180^{\circ}, 180^{\circ}]\): • For \(k = -2\): \(\theta = 22 - 144 = -122^{\circ}\). • For \(k = -1\): \(\theta = 22 - 72 = -50^{\circ}\). • For \(k = 0\): \(\theta = 22^{\circ}\). • For \(k = 1\): \(\theta = 22 + 72 = 94^{\circ}\). • For \(k = 2\): \(\theta = 22 + 144 = 166^{\circ}\). Checking \(k = -3\) gives \(-194^{\circ}\) (which is less than \(-180^{\circ}\)) and \(k = 3\) gives \(238^{\circ}\) (which is larger than \(180^{\circ}\)). Thus, the solutions are \[ \theta = -122^{\circ},\quad -50^{\circ},\quad 22^{\circ},\quad 94^{\circ},\quad 166^{\circ}. \] --- ### 2(b) We now need to find \(\theta \in [-180^{\circ}, 180^{\circ}]\) such that \[ 2 \cos 3\theta \cdot \cos 30^{\circ} - 2 \sin 3\theta \cdot \sin 30^{\circ} = 1. \] Factor out the 2: \[ 2 \left(\cos 3\theta \cdot \cos 30^{\circ} - \sin 3\theta \cdot \sin 30^{\circ}\right) = 1. \] Recall that \[ \cos 3\theta \cos 30^{\circ} - \sin 3\theta \sin 30^{\circ} = \cos(3\theta + 30^{\circ}). \] Thus, the equation becomes \[ 2 \cos(3\theta + 30^{\circ}) = 1, \] or \[ \cos(3\theta + 30^{\circ}) = \frac{1}{2}. \] Since \(\cos \alpha = \frac{1}{2}\) when \[ \alpha = 60^{\circ} + 360^{\circ} k \quad \text{or} \quad \alpha = 300^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z}, \] we set 1. \(3\theta + 30^{\circ} = 60^{\circ} + 360^{\circ} k\): \[ 3\theta = 30^{\circ} + 360^{\circ} k,\quad \theta = 10^{\circ} + 120^{\circ} k. \] 2. \(3\theta + 30^{\circ} = 300^{\circ} + 360^{\circ} k\): \[ 3\theta = 270^{\circ} + 360^{\circ} k,\quad \theta = 90^{\circ} + 120^{\circ} k. \] Now choose the values such that \(\theta \in [-180^{\circ}, 180^{\circ}]\). For \(\theta =