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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 \( x \in\left[180^{\circ} ; 360^{\circ}\right] \) \( 7.2 \quad \begin{array}{l}\text { Draw shetch graphs of } f \text { and } g \text { for } x \in\left[180^{\circ} ; 360^{\circ}\right] \text { on the same system of axes } \\ \text { provided on DIAGRAM SHEET } 2 \text {. } \\ 7.3\end{array} \begin{array}{l}\text { For which values of } x \text { will } f(x) \leq g(x) \text { for } x \in\left[180^{\circ} ; 360^{\circ}\right] ?\end{array} \)

Ask by O'Quinn Black. in South Africa
Mar 08,2025

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The graphs of \( f(x) = 1 + \sin x \) and \( g(x) = \cos(2x) \) intersect at \( x = 180^\circ, 210^\circ, 330^\circ, 360^\circ \) within the interval \( [180^\circ, 360^\circ] \). There are no values of \( x \) in this interval where \( f(x) \leq g(x) \) except at the points of intersection.

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To find the points of intersection between \(f(x) = 1 + \sin x\) and \(g(x) = \cos 2x\) for \(x \in [180^{\circ}; 360^{\circ}]\), we set the equations equal to each other: \[ 1 + \sin x = \cos 2x. \] Using the identity \(\cos 2x = 1 - 2\sin^2 x\), we can rewrite the equation as: \[ 1 + \sin x = 1 - 2\sin^2 x. \] \[ \sin x + 2\sin^2 x = 0. \] Factoring gives us \(\sin x(2\sin x + 1) = 0\), leading to two potential solutions: 1. \(\sin x = 0\) (which occurs at \(x = 180^{\circ}\) and \(x = 360^{\circ}\)), and 2. \(2\sin x + 1 = 0\) \(\Rightarrow \sin x = -\frac{1}{2}\) (which occurs at \(x = 210^{\circ}\) and \(x = 330^{\circ}\) in the specified range). Thus, the points of intersection in the given interval are at \(x = 180^{\circ}\), \(x = 210^{\circ}\), and \(x = 360^{\circ}\). --- When sketching the graphs of \(f\) and \(g\) on the interval \([180^{\circ}; 360^{\circ}]\), you'll find a sine wave-like function for \(f(x)\), peaking just above 2, while \(g(x)\) dips down, creating a cosine wave that hit values between -1 to 1. The intersection points can clearly be marked on the graph, showing how the functions interact and where \(f\) crosses \(g\). To determine where \(f(x) \leq g(x)\) within the interval, we can analyze the behavior of both functions. Since \(f(x)\) starts at 2 and decreases while \(g(x)\) oscillates between -1 and 1, the most significant interval where \(f(x) \leq g(x)\) is when \(x\) lies between the intersection points \(x = 210^{\circ}\) to \(x = 330^{\circ}\), where the sine graph dips below the cosine values.

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