tìm x thỏa: a) sinx = $\frac{1}{2}$ giúp vs coi 18/09/2021 Bởi Raelynn tìm x thỏa: a) sinx = $\frac{1}{2}$ giúp vs coi
sinx= $\frac{1}{2}$ ⇔sinx= sin $\frac{π}{6}$ ⇔\(\left[ \begin{array}{l}x=\frac{π}{6} +k2π\\x=π -\frac{π}{6}+k2π \end{array} \right.\) K∈Z ⇔\(\left[ \begin{array}{l}x=\frac{π}{6} +k2π\\x=\frac{5π}{6} +k2π\end{array} \right.\) K ∈Z Bình luận
đáp án: a) sinx = $\frac{1}{2}$ ⇔ sinx = sin$\frac{π}{6} ⇔ \(\left[ \begin{array}{l}x=\frac{π}{6}+k.2π \\x=π-\frac{π}{6}+k.2π \end{array} \right.\) ( k∈Z) ⇔ \(\left[ \begin{array}{l}x=\frac{π}{6} +k.2π\\x=5π/6+k.2π\end{array} \right.\) (k∈Z) Bình luận
sinx= $\frac{1}{2}$
⇔sinx= sin $\frac{π}{6}$
⇔\(\left[ \begin{array}{l}x=\frac{π}{6} +k2π\\x=π -\frac{π}{6}+k2π \end{array} \right.\) K∈Z
⇔\(\left[ \begin{array}{l}x=\frac{π}{6} +k2π\\x=\frac{5π}{6} +k2π\end{array} \right.\) K ∈Z
đáp án:
a) sinx = $\frac{1}{2}$
⇔ sinx = sin$\frac{π}{6}
⇔ \(\left[ \begin{array}{l}x=\frac{π}{6}+k.2π \\x=π-\frac{π}{6}+k.2π \end{array} \right.\) ( k∈Z)
⇔ \(\left[ \begin{array}{l}x=\frac{π}{6} +k.2π\\x=5π/6+k.2π\end{array} \right.\) (k∈Z)