Questions: Solve the equation on the interval (0 leq theta<2 pi). (cos (2 theta-fracpi2)=1)

Solution Steps

To solve the equation \(\cos \left(2 \theta-\frac{\pi}{2}\right)=1\) on the interval \(0 \leq \theta < 2 \pi\), we need to find the values of \(\theta\) that satisfy the equation. We can use the properties of the cosine function and its periodicity to determine these values.

1. Recognize that \(\cos(x) = 1\) when \(x = 2k\pi\) for any integer \(k\).
2. Set \(2 \theta - \frac{\pi}{2} = 2k\pi\) and solve for \(\theta\).
3. Determine the values of \(k\) that keep \(\theta\) within the interval \(0 \leq \theta < 2 \pi\).

We start with the equation

\[ \cos \left(2 \theta - \frac{\pi}{2}\right) = 1. \]

The cosine function equals 1 at angles of the form

\[ x = 2k\pi, \]

where \(k\) is any integer.

Setting

\[ 2 \theta - \frac{\pi}{2} = 2k\pi, \]

we can solve for \(\theta\):

\[ 2 \theta = 2k\pi + \frac{\pi}{2} \implies \theta = k\pi + \frac{\pi}{4}. \]

We need to find values of \(k\) such that

\[ 0 \leq \theta < 2\pi. \]

Calculating for different integer values of \(k\):

• For \(k = 0\): \[ \theta = 0 \cdot \pi + \frac{\pi}{4} = \frac{\pi}{4} \approx 0.7854. \]

• For \(k = 1\): \[ \theta = 1 \cdot \pi + \frac{\pi}{4} = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \approx 3.9269. \]

• For \(k = -1\): \[ \theta = -1 \cdot \pi + \frac{\pi}{4} = -\pi + \frac{\pi}{4} \text{ (not in the interval)}. \]

Thus, the valid solutions are

\[ \theta = \frac{\pi}{4} \text{ and } \frac{5\pi}{4}. \]

Final Answer