Given: \( g(x)=2 \cos \left(x-30^{\circ}\right) \) 4.1 Sketch the graph of \( g \) for \( x \in\left[-90^{\circ} ; 270^{\circ}\right] \) on DIAGRAM SHEET 1 . 4.2 Use the symbols \( A \) and \( B \) to plot the two points on the graph of \( g \) for whic \( \cos \left(x-30^{\circ}\right)=0.5 \) 4.3 Calculate the \( x \)-coordinates of the points A and B . 4.4 Write down the values of \( x \), where \( x \in\left[-90^{\circ} ; 270^{\circ}\right] \) and \( g^{\prime}(x)=0 \). 4.5 Use the graph to solve for \( x, x \in\left[-90^{\circ} ; 270^{\circ}\right] \), where \( g(x)<0 \)

Let's solve the problem step by step. ### 4.1 Sketch the graph of \( g(x) = 2 \cos(x - 30^{\circ}) \) To sketch the graph of \( g(x) \), we need to understand the properties of the cosine function and how the transformation affects it. 1. **Amplitude**: The amplitude of \( g(x) \) is 2, which means the graph will oscillate between -2 and 2. 2. **Period**: The period of the cosine function is \( 360^{\circ} \). Since there is no coefficient affecting \( x \), the period remains \( 360^{\circ} \). 3. **Phase Shift**: The function is shifted to the right by \( 30^{\circ} \). The key points for the cosine function are: - Maximum at \( 0^{\circ} \) - Zero at \( 90^{\circ} \) - Minimum at \( 180^{\circ} \) - Zero at \( 270^{\circ} \) - Maximum at \( 360^{\circ} \) Considering the phase shift, we can calculate the key points for \( g(x) \): - Maximum at \( 30^{\circ} \) (value = 2) - Zero at \( 120^{\circ} \) (value = 0) - Minimum at \( 210^{\circ} \) (value = -2) - Zero at \( 300^{\circ} \) (value = 0) ### 4.2 Plot the points \( A \) and \( B \) where \( \cos(x - 30^{\circ}) = 0.5 \) To find the points where \( \cos(x - 30^{\circ}) = 0.5 \), we can use the cosine inverse function: \[ x - 30^{\circ} = 60^{\circ} + k \cdot 360^{\circ} \quad \text{or} \quad x - 30^{\circ} = -60^{\circ} + k \cdot 360^{\circ} \] Solving for \( x \): 1. \( x - 30^{\circ} = 60^{\circ} \) gives \( x = 90^{\circ} \) 2. \( x - 30^{\circ} = -60^{\circ} \) gives \( x = -30^{\circ} \) Thus, the points \( A \) and \( B \) are: - \( A(90^{\circ}, g(90^{\circ})) \) - \( B(-30^{\circ}, g(-30^{\circ})) \) ### 4.3 Calculate the \( x \)-coordinates of the points \( A \) and \( B \) From the previous calculations: - Point \( A \): \( x_A = 90^{\circ} \) - Point \( B \): \( x_B = -30^{\circ} \) ### 4.4 Values of \( x \) where \( g'(x) = 0 \) To find where the derivative \( g'(x) = 0 \), we first need to calculate \( g'(x) \): \[ g'(x) = -2 \sin(x - 30^{\circ}) \] Setting the derivative to zero: \[ -2 \sin(x - 30^{\circ}) = 0 \implies \sin(x - 30^{\circ}) = 0 \] The sine function is zero at integer multiples of \( 180^{\circ} \): \[ x - 30^{\circ} = k \cdot 180^{\circ} \implies x = 30^{\circ} + k \cdot 180^{\circ} \] For \( k = 0 \): - \( x = 30^{\circ} \) For \( k = 1 \): - \( x = 210^{\circ} \) For \( k = -1 \): - \( x = -150^{\circ} \) (not in the interval) Thus, the values of \( x \) where \( g'(x) = 0 \) in the interval \( [-90^{\circ}, 270^{\circ}] \) are: - \( 30^{\circ} \) - \( 210^{\circ} \) ### 4.5 Solve for \( x \) where \( g(x) < 0 \) To find where \( g(x) < 0 \): - The function \( g(x) = 2 \cos(x - 30^{\circ}) < 0 \) when \( \cos(x - 30^{\circ}) < 0 \). The cosine function is negative in the intervals: - \( 90^{\circ} < x - 30^{\circ} < 270^{\circ} \) Solving for \( x \): 1. \( 90^{\circ} < x - 30^{\circ} \) gives \( x > 120^{\circ} \) 2. \( x - 30^{\circ} < 270^{\circ} \) gives \( x < 300^{\circ} \) Thus, the solution for \( g(x) < 0 \) in the interval \( [-90^{\circ}, 270^{\circ}] \) is: - \( x \in (120^{\circ}, 270^{\circ}) \) ### Summary of Results - **4.3**: \( A(90^{\circ}, g(90^{\circ})) \), \( B(-30^{\circ}, g(-30^{\circ})) \) - **4.4**: \( x = 30^{\circ}, 210^{\circ} \) - **4.5**: \( x \in (120^{\circ}, 270^{\circ}) \) where \( g(x) < 0 \)