solve(sin(x)=1/2)

We solve in

ℝ

\sin (x) = \frac{1}{2}

From sin(θ) =

\frac{1}{2}

, we deduce: θ =

\frac{π}{6} + 2 k π

or θ =

π - \frac{π}{6} + 2 k π

, k ∈ ℤ

We solve in

ℝ

x = π - \frac{π}{6} + 2 k π

ⓘ Group like terms by adding their coefficients:

π - \frac{π}{6}

\frac{5 π}{6}

Solution found:

x = \frac{5 π}{6} + 2 k π

Solutions obtained:

x = {\frac{π}{6} + 2 k π, \frac{5 π}{6} + 2 k π}

▶Solution:

{x = {\frac{π}{6} + 2 k π, \frac{5 π}{6} + 2 k π}, k \in ℤ}

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