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To solve for all angles \( \theta \) in the interval \([0, 2\pi)\) where \( \cos(\theta + \pi/6) = \sqrt{3}/2 \), we first identify the standard angles where cosine equals \( \sqrt{3}/2 \). These angles are:

$\alpha = 0 \text{ or } \alpha = 2\pi$

Next, we set up the equation:

$\theta + \pi/6 = 0 + 2k\pi \text{ or } \theta + \pi/6 = 2\pi + 2k\pi$

where \( k \) is an integer.

Solving these equations for \( \theta \):

\( \theta = -\pi/6 + 2k\pi \) or \( \theta = 11\pi/6 + 2k\pi \)

Since \( \theta \) must be in the interval \([0, 2\pi)\), we find specific solutions by setting \( k = 0 \):

\( \theta = -\pi/6 = 11\pi/6 \) (not in the interval)

and \( \theta = 11\pi/6 \text{ (valid)} $

We solve the equation ( cos( heta + pi/6) = sqrt{3}/2 ) by finding all standard angles where cosine equals ( sqrt{3}/2 ). These angles are:

$0 + 2kpi ext{ and } 2pi + 2kpi$

We set up the equation:

$ heta + pi/6 = 0 + 2kpi ext{ or } heta + pi/6 = 2pi + 2kpi$

Solving for ( heta ):

$ heta = -pi/6 + 2kpi ext{ or } heta = 11pi/6 + 2kpi $

Ensuring ( heta ) is within ([0, 2pi)), for ( k = 0 ):

$ heta = 11pi/6$

Solve ( cos( heta + pi/6) = sqrt{3}/2 ) by finding standard angles where cosine equals ( sqrt{3}/2 ):

$0 + 2kpi ext{ and } 2pi + 2kpi$

Set up:

$ heta + pi/6 = 0 + 2kpi ext{ or } heta + pi/6 = 2pi + 2kpi$

Solve:

$ heta = -pi/6 + 2kpi ext{ or } heta = 11pi/6 + 2kpi $

For ( k = 0 ):

$ heta = 11pi/6$

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