Solve for θ:
[tex]\[ 1 + \sin \theta = \frac{\sqrt{3} + 2}{2} \][/tex]

A. [tex]\(\frac{\pi}{3}, \frac{5\pi}{3}\)[/tex]

B. [tex]\(\frac{\pi}{6}, \frac{5\pi}{6}\)[/tex]

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

D. [tex]\(\frac{2\pi}{3}, \frac{4\pi}{3}\)[/tex]



Answer :

To solve the equation [tex]\( 1 + \sin \theta = \frac{\sqrt{3} + 2}{2} \)[/tex], we'll proceed step-by-step.

1. Isolate [tex]\(\sin \theta\)[/tex]:

Subtract 1 from both sides of the equation:
[tex]\[ \sin \theta = \frac{\sqrt{3} + 2}{2} - 1 \][/tex]

2. Simplify the right-hand side:

Simplify the expression on the right-hand side:
[tex]\[ \sin \theta = \frac{\sqrt{3} + 2}{2} - \frac{2}{2} = \frac{\sqrt{3} + 2 - 2}{2} = \frac{\sqrt{3}}{2} \][/tex]

3. Set the equation for [tex]\(\sin \theta\)[/tex]:

Now we need to solve:
[tex]\[ \sin \theta = \frac{\sqrt{3}}{2} \][/tex]

4. Find the angles that satisfy the equation:

We know from trigonometry that [tex]\(\sin \theta = \frac{\sqrt{3}}{2}\)[/tex] is true for:
[tex]\(\theta = \frac{\pi}{3}\)[/tex] and [tex]\(\theta = \frac{2\pi}{3} + 2k\pi\)[/tex]

However, we need to find solutions in the interval [tex]\( [0, 2\pi] \)[/tex]. Observing the symmetry and values on the unit circle:

[tex]\[ \theta = \frac{\pi}{3}, \quad \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \][/tex]
[tex]\[ \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \][/tex]

5. Verify the solutions and check intervals:

In the interval [tex]\( [0, 2\pi] \)[/tex], the solutions are:
[tex]\(\theta = \frac{\pi}{3}\)[/tex] and [tex]\(\theta = \frac{5\pi}{3}\)[/tex]

Thus, the correct answer is:
A. [tex]\(\frac{\pi}{3}, \frac{5 \pi}{3}\)[/tex]