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