Answer :
Certainly! Let's solve the equation step by step:
Step 1: Start with the given equation.
[tex]\[ 4\cos^2(x) = 3 \][/tex]
Step 2: Isolate [tex]\(\cos^2(x)\)[/tex].
Divide both sides of the equation by 4:
[tex]\[ \cos^2(x) = \frac{3}{4} \][/tex]
Step 3: Take the square root of both sides.
[tex]\[ \cos(x) = \pm \sqrt{\frac{3}{4}} \][/tex]
Simplify the square root expression:
[tex]\[ \cos(x) = \pm \frac{\sqrt{3}}{2} \][/tex]
Here, [tex]\(\cos(x)\)[/tex] can take on two values:
[tex]\[ \cos(x) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos(x) = -\frac{\sqrt{3}}{2} \][/tex]
Step 4: Solve for [tex]\(x\)[/tex] using the inverse cosine function (arc cosine).
For [tex]\(\cos(x) = \frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ x = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ x = 0.5235987755982989 \][/tex]
Since cosine is positive in the first and fourth quadrants, the other solution within [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex] interval is:
[tex]\[ x = 2\pi - 0.5235987755982989 \][/tex]
[tex]\[ x = 5.759586531581287 \][/tex]
For [tex]\(\cos(x) = -\frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ x = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ x = 2.6179938779914944 \][/tex]
Since cosine is negative in the second and third quadrants, the other solution within [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex] interval is:
[tex]\[ x = 2\pi - 2.6179938779914944 \][/tex]
[tex]\[ x = 3.665191429188092 \][/tex]
Step 5: Write the general solutions considering the periodicity of the cosine function.
The cosine function has a period of [tex]\(2\pi\)[/tex], so we add multiples of [tex]\(2\pi\)[/tex] (denoted as [tex]\(2k\pi\)[/tex], where [tex]\(k\)[/tex] is any integer) to each solution:
Therefore, the general solutions are:
[tex]\[ x = 0.5235987755982989 + 2k\pi \][/tex]
[tex]\[ x = 5.759586531581287 + 2k\pi \][/tex]
[tex]\[ x = 2.6179938779914944 + 2k\pi \][/tex]
[tex]\[ x = 3.665191429188092 + 2k\pi \][/tex]
Where [tex]\(k \in \mathbb{Z}\)[/tex] (k is any integer).
Thus, the x-intercepts of the graph [tex]\(4\cos^2(x) = 3\)[/tex] are given by these general solutions.
Step 1: Start with the given equation.
[tex]\[ 4\cos^2(x) = 3 \][/tex]
Step 2: Isolate [tex]\(\cos^2(x)\)[/tex].
Divide both sides of the equation by 4:
[tex]\[ \cos^2(x) = \frac{3}{4} \][/tex]
Step 3: Take the square root of both sides.
[tex]\[ \cos(x) = \pm \sqrt{\frac{3}{4}} \][/tex]
Simplify the square root expression:
[tex]\[ \cos(x) = \pm \frac{\sqrt{3}}{2} \][/tex]
Here, [tex]\(\cos(x)\)[/tex] can take on two values:
[tex]\[ \cos(x) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos(x) = -\frac{\sqrt{3}}{2} \][/tex]
Step 4: Solve for [tex]\(x\)[/tex] using the inverse cosine function (arc cosine).
For [tex]\(\cos(x) = \frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ x = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ x = 0.5235987755982989 \][/tex]
Since cosine is positive in the first and fourth quadrants, the other solution within [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex] interval is:
[tex]\[ x = 2\pi - 0.5235987755982989 \][/tex]
[tex]\[ x = 5.759586531581287 \][/tex]
For [tex]\(\cos(x) = -\frac{\sqrt{3}}{2}\)[/tex]:
[tex]\[ x = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \][/tex]
[tex]\[ x = 2.6179938779914944 \][/tex]
Since cosine is negative in the second and third quadrants, the other solution within [tex]\(0\)[/tex] to [tex]\(2\pi\)[/tex] interval is:
[tex]\[ x = 2\pi - 2.6179938779914944 \][/tex]
[tex]\[ x = 3.665191429188092 \][/tex]
Step 5: Write the general solutions considering the periodicity of the cosine function.
The cosine function has a period of [tex]\(2\pi\)[/tex], so we add multiples of [tex]\(2\pi\)[/tex] (denoted as [tex]\(2k\pi\)[/tex], where [tex]\(k\)[/tex] is any integer) to each solution:
Therefore, the general solutions are:
[tex]\[ x = 0.5235987755982989 + 2k\pi \][/tex]
[tex]\[ x = 5.759586531581287 + 2k\pi \][/tex]
[tex]\[ x = 2.6179938779914944 + 2k\pi \][/tex]
[tex]\[ x = 3.665191429188092 + 2k\pi \][/tex]
Where [tex]\(k \in \mathbb{Z}\)[/tex] (k is any integer).
Thus, the x-intercepts of the graph [tex]\(4\cos^2(x) = 3\)[/tex] are given by these general solutions.
