To find the real zeros of the trigonometric function [tex]\( f(x) = \cos(2x) + 1 \)[/tex] within the interval [tex]\( 0 \leq \theta < 2\pi \)[/tex], we need to solve for [tex]\( x \)[/tex] such that [tex]\( f(x) = 0 \)[/tex].
1. Set the function equal to zero:
[tex]\[
\cos(2x) + 1 = 0
\][/tex]
2. Isolate the cosine term:
[tex]\[
\cos(2x) = -1
\][/tex]
3. Determine the angles where the cosine function equals -1:
The cosine function equals -1 at:
[tex]\[
2x = \pi + 2k\pi \quad \text{for integer } k
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
Since [tex]\( \frac{\pi}{2} \)[/tex] is the smallest positive solution and repeating every [tex]\(\pi \)[/tex], we should find solutions within the specified interval [tex]\( 0 \leq x < 2\pi \)[/tex].
[tex]\[
2x = \pi \implies x = \frac{\pi}{2}
\][/tex]
Another solution occurs one full cycle later:
[tex]\[
2x = 3\pi \implies x = \frac{3\pi}{2}
\][/tex]
5. Confirm that the solutions are within the given interval:
Both [tex]\( x = \frac{\pi}{2} \)[/tex] and [tex]\( x = \frac{3\pi}{2} \)[/tex] lie within the interval [tex]\( 0 \leq x < 2\pi \)[/tex].
Thus, the real zeros of the function [tex]\( f(x) = \cos(2x) + 1 \)[/tex] in the interval [tex]\( 0 \leq \theta < 2\pi \)[/tex] are:
[tex]\[
x = \frac{\pi}{2} \quad \text{ and } \quad x = \frac{3\pi}{2}
\][/tex]
To convert these solutions into their numerical form, we have:
[tex]\[
x_1 = \frac{\pi}{2} \approx 1.5707963267948966
\][/tex]
[tex]\[
x_2 = \frac{3\pi}{2} \approx 4.71238898038469
\][/tex]
Therefore, the values of [tex]\( x \)[/tex] are approximately:
[tex]\[
x \approx 1.5707963267948966 \quad \text{and} \quad x \approx 4.71238898038469
\][/tex]
These are the zeros of the function [tex]\( f(x) = \cos(2x) + 1 \)[/tex] in the interval [tex]\( 0 \leq \theta < 2\pi \)[/tex].
