`sin (2x - 10^0) = 1/2`
`<=> sin (2x - 10^0) = sin 30^0`
`<=>` \(\left[ \begin{array}{l}2x - 10^0 = 30^0 + k360^0\\2x - 10^0 = 150^0 + k360^0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x = 40^0 + k360^0\\2x = 160^0 + k360^0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = 20^0 + k180^0\\x = 80^0 + k180^0\end{array} \right.\) `(k ∈ ZZ)`
Vậy
\(\left[ \begin{array}{l}-120^0 < 20^0 + k180^0 < 90^0\\-120^0 < 80^0 + k180^0 < 90^0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}-140^0 < k180^0 < 70^0\\-200^0 < k180^0 < 10^0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}-\dfrac{7}{9} < k < \dfrac{7}{18}\\-\dfrac{10}{9} < k < \dfrac{1}{18}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}-0,8 < k < 0,39\\-1,11 < k < 0,056\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}k ∈ {-1; 0}\\k ∈ {0}\end{array} \right.\)
Vậy có 2 giá trị `k` là` {-1; 0}`
