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At what [tex]$x$[/tex]-values do the graphs of the functions [tex]$y=\cos 2x$[/tex] and [tex]$y=\cos^2 x - 1$[/tex] intersect over the interval [tex]$0 \leq x \ \textless \ 2\pi$[/tex]? Select all that apply.

A. [tex]$-\frac{\pi}{2}$[/tex]
B. [tex]$\frac{\pi}{2}$[/tex]
C. [tex]$\pi$[/tex]
D. [tex]$\frac{3\pi}{2}$[/tex]


Sagot :

To determine the [tex]\(x\)[/tex]-values at which the graphs of the functions [tex]\(y = \cos(2x)\)[/tex] and [tex]\(y = \cos^2(x) - 1\)[/tex] intersect over the interval [tex]\(0 \leq x < 2\pi\)[/tex], we need to set the equations equal to each other and solve for [tex]\(x\)[/tex].

First, let's set the two functions equal to each other:

[tex]\[ \cos(2x) = \cos^2(x) - 1 \][/tex]

Next, let's use a trigonometric identity for [tex]\(\cos(2x)\)[/tex]:

[tex]\[ \cos(2x) = 2\cos^2(x) - 1 \][/tex]

Now, we can rewrite the equation substituting [tex]\(\cos(2x)\)[/tex]:

[tex]\[ 2\cos^2(x) - 1 = \cos^2(x) - 1 \][/tex]

To simplify this, we subtract [tex]\((\cos^2(x) - 1)\)[/tex] from both sides:

[tex]\[ 2\cos^2(x) - 1 - \cos^2(x) + 1 = 0 \][/tex]

This simplifies to:

[tex]\[ \cos^2(x) = 0 \][/tex]

Taking the square root of both sides, we get:

[tex]\[ \cos(x) = 0 \][/tex]

Now, we need to find the values of [tex]\(x\)[/tex] within the given interval [tex]\(0 \leq x < 2\pi\)[/tex] where [tex]\(\cos(x) = 0\)[/tex]. These values occur at:

[tex]\[ x = \frac{\pi}{2} \quad \text{and} \quad x = \frac{3\pi}{2} \][/tex]

Both of these [tex]\(x\)[/tex]-values are within the interval [tex]\(0 \leq x < 2\pi\)[/tex], and thus, they are the points at which the graphs of the functions intersect.

Hence, the correct [tex]\(x\)[/tex]-values where the graphs of [tex]\(y = \cos(2x)\)[/tex] and [tex]\(y = \cos^2(x) - 1\)[/tex] intersect are:

[tex]\[ \boxed{\frac{\pi}{2}, \frac{3\pi}{2}} \][/tex]

The options you should select are:

[tex]\(\frac{\pi}{2}\)[/tex]

[tex]\(\frac{3\pi}{2}\)[/tex]