To find all solutions of the equation [tex]\( 2 \sin \theta - \sqrt{3} = 0 \)[/tex] in radians, follow these steps:
1. Isolate [tex]\(\sin \theta\)[/tex]:
[tex]\[
2 \sin \theta = \sqrt{3}
\][/tex]
[tex]\[
\sin \theta = \frac{\sqrt{3}}{2}
\][/tex]
2. Determine the general solutions for [tex]\(\sin \theta = \frac{\sqrt{3}}{2}\)[/tex]. The angles that satisfy this equation within one period [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
\theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{2\pi}{3}
\][/tex]
3. Include all possible solutions by adding integer multiples of [tex]\( 2\pi \)[/tex] to the general solutions:
[tex]\[
\theta = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad \theta = \frac{2\pi}{3} + 2k\pi \quad \text{for integer} \, k
\][/tex]
4. Represent these solutions within one period [tex]\([0, 2\pi)\)[/tex]:
By reducing these expressions modulo [tex]\( 2\pi \)[/tex], each unique angle within one period is considered.
From analyzing the result, the solutions within one period are:
[tex]\[
\theta = 1.0471975511965965, 2.094395102393193, 2.0943951023931966, 2.094395102393195, 1.0471975511965983, 1.0471975511965974, 1.0471975511965976, 2.0943951023931953
\][/tex]
Given these repetitive values due to minor numerical variances, we can simplify to two principal solutions:
[tex]\[
\theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{2\pi}{3}
\][/tex]
Thus, the final answers for [tex]\(\theta\)[/tex] in radians, within the given period, are:
[tex]\[
\theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{2\pi}{3}
\][/tex]
