Đáp án:
`a, S= {π/3+k2π; (2π)/3 +k2π, k \in ZZ}`
`b, S= { arc sin (1/4) + k2π; π -arc sin (1/4) + k2π, k \in ZZ}`
`c, S ={-(5π)/(12) + k2π; (7π)/(12)+k2π ,k\in ZZ}`
Giải thích các bước giải:
`a, sinx = (\sqrt{3})/2`
`<=> sin x = sin \frac{π}{3}`
`<=>` \(\left[ \begin{array}{l}x= \dfracπ3 + k2π \\x=π - \dfracπ3 +k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x= \dfracπ3 + k2π \\x= \dfrac{2π}{3} +k2π\end{array} \right.\) `(k \in ZZ)`
Vậy `S= {π/3+k2π; (2π)/3 +k2π, k \in ZZ}`
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`b, sin x = 1/4`
`<=>` \(\left[ \begin{array}{l}x=arc sin (\dfrac{1}{4}) +k2π \\x=π - arcsin (\dfrac{1}{4}) + k2π\end{array} \right.\)
Vậy ` S= { arc sin (1/4) + k2π; π -arc sin (1/4) + k2π, k \in ZZ}`
_____________
`c, sin (x +30°) = - (\sqrt{2})/2`
`<=> sin (x +π/6) = sin \frac{-π}{4}`
`<=>` \(\left[ \begin{array}{l}x+\dfracπ6 = -\dfracπ4+ k2π\\x+\dfracπ6 = π-\dfracπ4 +k2π\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = -\dfrac{5π}{12}+ k2π\\x= \dfrac{7π}{12} +k2π\end{array} \right.\)
Vậy `S ={-(5π)/(12) + k2π; (7π)/(12)+k2π ,k\in ZZ}`
