3.1.2 4. Calculate \( x \) if \( 2 \sin \left(x-30^{\circ}\right)=\cot 150^{\circ} a n d 0^{\circ} \leq x \leq 360^{\circ} \) 3.1 \( \cos \left(\phi+10^{\circ}\right)=\frac{1}{2} ; \) if \( 0^{\circ}<\phi+10^{\circ}<10^{\circ}<360^{\circ} \) 3.2 Sketch the graphs \( y=\tan x \) and \( y=2 \sin x \) on the same set of axes for the domain \( 10^{\circ} ; 360^{\circ} \). Show clearly on your sketch how to Obtain the values of \( x \) where: 3.2.1 \( \frac{1}{2} \tan x=\sin x \)

Let's solve each part of your question step by step. ### Part 4: Solve for \( x \) We start with the equation: \[ 2 \sin \left(x - 30^{\circ}\right) = \cot 150^{\circ} \] First, we need to find \( \cot 150^{\circ} \). We know that: \[ \cot 150^{\circ} = \frac{1}{\tan 150^{\circ}} = \frac{1}{-\tan 30^{\circ}} = -\frac{1}{\frac{1}{\sqrt{3}}} = -\sqrt{3} \] Now, substituting this back into the equation gives us: \[ 2 \sin \left(x - 30^{\circ}\right) = -\sqrt{3} \] Dividing both sides by 2: \[ \sin \left(x - 30^{\circ}\right) = -\frac{\sqrt{3}}{2} \] The sine function equals \(-\frac{\sqrt{3}}{2}\) at the angles: \[ x - 30^{\circ} = 240^{\circ} + k \cdot 360^{\circ} \quad \text{and} \quad x - 30^{\circ} = 300^{\circ} + k \cdot 360^{\circ} \] for \( k \in \mathbb{Z} \). Now, solving for \( x \): 1. For \( x - 30^{\circ} = 240^{\circ} \): \[ x = 240^{\circ} + 30^{\circ} = 270^{\circ} \] 2. For \( x - 30^{\circ} = 300^{\circ} \): \[ x = 300^{\circ} + 30^{\circ} = 330^{\circ} \] Now, we check if these values are within the range \( 0^{\circ} \leq x \leq 360^{\circ} \): - \( x = 270^{\circ} \) is valid. - \( x = 330^{\circ} \) is valid. Thus, the solutions for \( x \) are: \[ \boxed{270^{\circ}, 330^{\circ}} \] ### Part 3.1: Solve for \( \phi \) We have the equation: \[ \cos \left(\phi + 10^{\circ}\right) = \frac{1}{2} \] The cosine function equals \(\frac{1}{2}\) at: \[ \phi + 10^{\circ} = 60^{\circ} + k \cdot 360^{\circ} \quad \text{and} \quad \phi + 10^{\circ} = 300^{\circ} + k \cdot 360^{\circ} \] for \( k \in \mathbb{Z} \). Now, solving for \( \phi \): 1. For \( \phi + 10^{\circ} = 60^{\circ} \): \[ \phi = 60^{\circ} - 10^{\circ} = 50^{\circ} \] 2. For \( \phi + 10^{\circ} = 300^{\circ} \): \[ \phi = 300^{\circ} - 10^{\circ} = 290^{\circ} \] Now, we check if these values are within the range \( 0^{\circ} < \phi + 10^{\circ} < 360^{\circ} \): - \( \phi = 50^{\circ} \) is valid. - \( \phi = 290^{\circ} \) is valid. Thus, the solutions for \( \phi \) are: \[ \boxed{50^{\circ}, 290^{\circ}} \] ### Part 3.2: Sketch the graphs To sketch the graphs of \( y = \tan x \) and \( y = 2 \sin x \) for the domain \( 10^{\circ} \leq x \leq 360^{\circ} \): 1. **Graph of \( y = \tan x \)**: - The tangent function has vertical asymptotes at \( x = 90^{\circ} \) and \( x = 270^{\circ} \). - It crosses the x-axis at \( x = 0^{\circ}, 180^{\circ}, 360^{\circ} \). 2. **