10) \( 3 \cos \left(2 x-\frac{\pi}{3}\right)-1=0 \)

Решить уравнение \(3 \cos \left(2 x-\frac{\pi}{3}\right)-1=0\). Solve the equation by following steps: - step0: Solve for \(x\): \(3\cos\left(2x-\frac{\pi }{3}\right)-1=0\) - step1: Use the periodicity identities: \(3\cos\left(2x+\frac{5\pi }{3}\right)-1=0\) - step2: Move the constant to the right side: \(3\cos\left(2x+\frac{5\pi }{3}\right)=0+1\) - step3: Remove 0: \(3\cos\left(2x+\frac{5\pi }{3}\right)=1\) - step4: Divide both sides: \(\frac{3\cos\left(2x+\frac{5\pi }{3}\right)}{3}=\frac{1}{3}\) - step5: Divide the numbers: \(\cos\left(2x+\frac{5\pi }{3}\right)=\frac{1}{3}\) - step6: Use the inverse trigonometric function: \(2x+\frac{5\pi }{3}=\arccos\left(\frac{1}{3}\right)\) - step7: Calculate: \(\begin{align}&2x+\frac{5\pi }{3}=-\arccos\left(\frac{1}{3}\right)\\&2x+\frac{5\pi }{3}=\arccos\left(\frac{1}{3}\right)\end{align}\) - step8: Add the period: \(\begin{align}&2x+\frac{5\pi }{3}=-\arccos\left(\frac{1}{3}\right)+2k\pi ,k \in \mathbb{Z}\\&2x+\frac{5\pi }{3}=\arccos\left(\frac{1}{3}\right)+2k\pi ,k \in \mathbb{Z}\end{align}\) - step9: Calculate: \(\begin{align}&x=-\frac{3\arccos\left(\frac{1}{3}\right)+5\pi }{6}+k\pi ,k \in \mathbb{Z}\\&2x+\frac{5\pi }{3}=\arccos\left(\frac{1}{3}\right)+2k\pi ,k \in \mathbb{Z}\end{align}\) - step10: Calculate: \(\begin{align}&x=-\frac{3\arccos\left(\frac{1}{3}\right)+5\pi }{6}+k\pi ,k \in \mathbb{Z}\\&x=\frac{3\arccos\left(\frac{1}{3}\right)-5\pi }{6}+k\pi ,k \in \mathbb{Z}\end{align}\) - step11: Find the union: \(x=\left\{ \begin{array}{l}-\frac{3\arccos\left(\frac{1}{3}\right)+5\pi }{6}+k\pi \\\frac{3\arccos\left(\frac{1}{3}\right)-5\pi }{6}+k\pi \end{array}\right.,k \in \mathbb{Z}\) Решение уравнения \(3 \cos \left(2 x-\frac{\pi}{3}\right)-1=0\) имеет вид: \[x = \left\{ \begin{array}{l} -\frac{3\arccos(\frac{1}{3})+5\pi}{6}+k\pi \\ \frac{3\arccos(\frac{1}{3})-5\pi}{6}+k\pi \end{array} \right., k \in \mathbb{Z}\] где \(k\) — целое число.