QUESTION 7 Given: \( f(x)=1+\sin x \) and \( g(x)=\cos 2 x \) \( 7.1 \quad \) Calculate the points of intersection of the graphs \( f \) and \( g \) for \( x \in\left[180^{\circ} ; 360^{\circ}\right] \) \( 7.2 \quad \begin{array}{l}\text { Draw shetch graphs of } f \text { and } g \text { for } x \in\left[180^{\circ} ; 360^{\circ}\right] \text { on the same system of axes } \\ \text { provided on DIAGRAM SHEET } 2 \text {. } \\ 7.3\end{array} \begin{array}{l}\text { For which values of } x \text { will } f(x) \leq g(x) \text { for } x \in\left[180^{\circ} ; 360^{\circ}\right] ?\end{array} \)

To solve the problem, we will follow the steps outlined in the question. ### Step 1: Calculate the points of intersection of the graphs \( f \) and \( g \) We need to find the values of \( x \) where \( f(x) = g(x) \). This means we need to solve the equation: \[ 1 + \sin x = \cos(2x) \] ### Step 2: Rewrite the equation Using the double angle identity for cosine, we can rewrite \( \cos(2x) \) as: \[ \cos(2x) = 1 - 2\sin^2(x) \] Substituting this into our equation gives: \[ 1 + \sin x = 1 - 2\sin^2(x) \] ### Step 3: Simplify the equation Subtracting 1 from both sides: \[ \sin x = -2\sin^2(x) \] Rearranging gives: \[ 2\sin^2(x) + \sin x = 0 \] ### Step 4: Factor the equation Factoring out \( \sin x \): \[ \sin x (2\sin x + 1) = 0 \] This gives us two cases to solve: 1. \( \sin x = 0 \) 2. \( 2\sin x + 1 = 0 \) or \( \sin x = -\frac{1}{2} \) ### Step 5: Solve for \( x \) 1. **For \( \sin x = 0 \)**: - The solutions in the interval \( [180^\circ, 360^\circ] \) are \( x = 180^\circ \) and \( x = 360^\circ \). 2. **For \( \sin x = -\frac{1}{2} \)**: - The solutions in the interval \( [180^\circ, 360^\circ] \) are \( x = 210^\circ \) and \( x = 330^\circ \). ### Step 6: Points of intersection Thus, the points of intersection of the graphs \( f \) and \( g \) in the interval \( [180^\circ, 360^\circ] \) are: \[ x = 180^\circ, \quad x = 210^\circ, \quad x = 330^\circ, \quad x = 360^\circ \] ### Step 7: Sketch the graphs For sketching the graphs of \( f(x) = 1 + \sin x \) and \( g(x) = \cos(2x) \) in the interval \( [180^\circ, 360^\circ] \): - \( f(x) \) will oscillate between 1 and 2. - \( g(x) \) will oscillate between -1 and 1. ### Step 8: Determine where \( f(x) \leq g(x) \) To find where \( f(x) \leq g(x) \): We need to solve the inequality: \[ 1 + \sin x \leq \cos(2x) \] This can be rewritten as: \[ \sin x + 1 \leq 1 - 2\sin^2(x) \] Simplifying gives: \[ 2\sin^2(x) + \sin x \leq 0 \] The solutions to this inequality will be the same as the points of intersection we found earlier. We will check the intervals between the points of intersection to determine where the inequality holds. ### Final Answer 1. Points of intersection: \( x = 180^\circ, 210^\circ, 330^\circ, 360^\circ \) 2. For \( f(x) \leq g(x) \), we will check the intervals: - Between \( 180^\circ \) and \( 210^\circ \) - Between \( 210^\circ \) and \( 330^\circ \) - Between \( 330^\circ \) and \( 360^\circ \) Let's evaluate the intervals to find where \( f(x) \leq g(x) \). We will check the values of \( f(x) \) and \( g(x) \) at key points in these intervals. - For \( x = 180^\circ \): \( f(180^\circ) = 1 + 0 = 1 \), \( g(180^\circ) = 1 \) → \( f(180^\circ) = g(180^\circ) \) - For \( x = 210^\circ \): \( f(210^\circ) = 1 - \frac{1}{2} = \frac{1}{2} \), \( g(210^\circ) = -\frac{1}{2} \) → \( f(210^\circ) > g(210^\circ) \) - For \( x = 330^\circ \): \( f(330^\circ) = 1 - \frac{1}{2} = \frac{1}{2} \), \( g(330^\circ) = -\frac{1}{2} \) → \( f(330^\circ) > g(330^\circ) \) - For \( x = 360^\circ \): \( f(360^\circ) = 1 + 0 = 1 \), \( g(360^\circ) = 1 \) → \( f(360^\circ) = g(360^\circ) \) Thus, \( f(x) \leq g(x) \) does not hold in the intervals between the points of intersection except at the points themselves. ### Conclusion - \( f(x) \leq g(x) \) does not hold for any \( x \) in \( [180^\circ, 360^\circ] \) except at the points of intersection.