chapter 7 quiz\n150 points possible answered: 8/15\nquestion 9\nfind all solutions of the equation ( 2 cos 3…

Explanation:

Step1: Solve for (\cos3x)

Divide both sides of (2\cos3x = 1) by (2): (\cos3x=\frac{1}{2})

Step2: Find general solutions for (3x)

We know that if (\cos\theta=\frac{1}{2}), then (\theta = 2k\pi\pm\frac{\pi}{3},k\in\mathbb{Z}). So for (\theta = 3x), we have (3x=2k\pi\pm\frac{\pi}{3}) Then (x=\frac{2k\pi}{3}\pm\frac{\pi}{9})

Step3: Find solutions in the interval ([0,\pi))

• When (k = 0): (x=\frac{\pi}{9}) (using the (+) sign in (x=\frac{2k\pi}{3}\pm\frac{\pi}{9}))
• When (k = 1): First, using the (+) sign: (x=\frac{2\pi}{3}+\frac{\pi}{9}=\frac{6\pi + \pi}{9}=\frac{7\pi}{9}) Second, using the (-) sign: (x=\frac{2\pi}{3}-\frac{\pi}{9}=\frac{6\pi-\pi}{9}=\frac{5\pi}{9})
• When (k = 2): (x=\frac{4\pi}{3}\pm\frac{\pi}{9}). For (x=\frac{4\pi}{3}+\frac{\pi}{9}=\frac{12\pi+\pi}{9}=\frac{13\pi}{9}\gt\pi) and (x=\frac{4\pi}{3}-\frac{\pi}{9}=\frac{12\pi - \pi}{9}=\frac{11\pi}{9}\gt\pi)

Answer:

(x_1=\frac{\pi}{9},x_2=\frac{5\pi}{9},x_3=\frac{7\pi}{9})