To solve the equation [tex]\(-3 \tan^2(x) + 1 = 0\)[/tex], let’s go through the process step by step.
1. Isolate [tex]\(\tan^2(x)\)[/tex]:
[tex]\[
-3 \tan^2(x) + 1 = 0
\][/tex]
[tex]\[
-3 \tan^2(x) = -1
\][/tex]
[tex]\[
\tan^2(x) = \frac{1}{3}
\][/tex]
2. Take the square root on both sides:
[tex]\[
\tan(x) = \pm \sqrt{\frac{1}{3}}
\][/tex]
[tex]\[
\tan(x) = \pm \frac{1}{\sqrt{3}}
\][/tex]
Simplifying, we get:
[tex]\[
\tan(x) = \pm \frac{\sqrt{3}}{3}
\][/tex]
3. Identify the angles whose tangent values are [tex]\(\frac{\sqrt{3}}{3}\)[/tex] and [tex]\(-\frac{\sqrt{3}}{3}\)[/tex]:
[tex]\(\tan(x) = \frac{\sqrt{3}}{3}\)[/tex] occurs at [tex]\(x = \frac{\pi}{6} + k\pi\)[/tex], where [tex]\(k\)[/tex] is any integer.
[tex]\(\tan(x) = -\frac{\sqrt{3}}{3}\)[/tex] occurs at [tex]\(x = \frac{5\pi}{6} + k\pi\)[/tex], where [tex]\(k\)[/tex] is any integer.
4. Combine the two sets of solutions:
Therefore, the complete set of solutions to the equation [tex]\(-3 \tan^2(x) + 1 = 0\)[/tex] in radians is:
[tex]\[
x = \frac{\pi}{6} + k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + k\pi \quad \text{for integer } k
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\frac{\pi}{6} + k\pi \quad \text{and} \quad \frac{5\pi}{6} + k\pi}
\][/tex]
