Determine the value of \( x \) to one decimal place \( \cos 2 x=0,50 \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\cos\left(2x\right)=0.5\) - step1: Use the inverse trigonometric function: \(2x=\arccos\left(0.5\right)\) - step2: Calculate: \(\begin{align}&2x=\frac{\pi }{3}\\&2x=\frac{5\pi }{3}\end{align}\) - step3: Add the period: \(\begin{align}&2x=\frac{\pi }{3}+2k\pi ,k \in \mathbb{Z}\\&2x=\frac{5\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step4: Calculate: \(\begin{align}&x=\frac{\pi }{6}+k\pi ,k \in \mathbb{Z}\\&2x=\frac{5\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step5: Calculate: \(\begin{align}&x=\frac{\pi }{6}+k\pi ,k \in \mathbb{Z}\\&x=\frac{5\pi }{6}+k\pi ,k \in \mathbb{Z}\end{align}\) - step6: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi }{6}+k\pi \\\frac{5\pi }{6}+k\pi \end{array}\right.,k \in \mathbb{Z}\) To solve the equation \( \cos 2x = 0.5 \), we can find the general solutions for \( 2x \) first. The cosine function equals \( 0.5 \) at the angles: - \( 2x = \frac{\pi}{3} + 2k\pi \) - \( 2x = \frac{5\pi}{3} + 2k\pi \) where \( k \) is any integer. Now, we can solve for \( x \) by dividing each equation by 2: 1. From \( 2x = \frac{\pi}{3} + 2k\pi \): \[ x = \frac{\pi}{6} + k\pi \] 2. From \( 2x = \frac{5\pi}{3} + 2k\pi \): \[ x = \frac{5\pi}{6} + k\pi \] Next, we can find the approximate values of \( x \) for \( k = 0 \): 1. For \( k = 0 \): - \( x = \frac{\pi}{6} \approx 0.524 \) - \( x = \frac{5\pi}{6} \approx 2.618 \) Now, rounding these values to one decimal place, we have: - \( x \approx 0.5 \) - \( x \approx 2.6 \) Thus, the values of \( x \) to one decimal place are approximately \( 0.5 \) and \( 2.6 \).