4. maximum mark: 5\nfind all points where the tangent to the curve ( y = 2 cos x - cos 2x ), ( 0 leq x leq…

Explanation:

Step1: Find the derivative of the function

The derivative of (y = 2\cos x-\cos2x) is (y^\prime=-2\sin x + 2\sin2x). Using the double - angle formula (\sin2x = 2\sin x\cos x), we get (y^\prime=-2\sin x+4\sin x\cos x=2\sin x(2\cos x - 1)).

Step2: Set the derivative equal to zero

Since the tangent is parallel to the (x) - axis when (y^\prime = 0), we set (2\sin x(2\cos x - 1)=0).

• Case 1: (\sin x=0). For (0\leq x\leq2\pi), (x = 0,\pi,2\pi).
• Case 2: (2\cos x-1 = 0), then (\cos x=\frac{1}{2}). For (0\leq x\leq2\pi), (x=\frac{\pi}{3},\frac{5\pi}{3}).

Step3: Find the (y) - values

• When (x = 0), (y=2\cos0-\cos0=2 - 1=1).
• When (x=\frac{\pi}{3}), (y = 2\cos\frac{\pi}{3}-\cos\frac{2\pi}{3}=2\times\frac{1}{2}-\left(-\frac{1}{2}\right)=\frac{3}{2}).
• When (x=\pi), (y=2\cos\pi-\cos2\pi=-2 - 1=-3).
• When (x=\frac{5\pi}{3}), (y = 2\cos\frac{5\pi}{3}-\cos\frac{10\pi}{3}=2\times\frac{1}{2}-\left(-\frac{1}{2}\right)=\frac{3}{2}).
• When (x = 2\pi), (y=2\cos2\pi-\cos4\pi=2 - 1=1).

Answer:

The points are ((0,1),\left(\frac{\pi}{3},\frac{3}{2}\right),(\pi, - 3),\left(\frac{5\pi}{3},\frac{3}{2}\right),(2\pi,1))