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(x+ 30 độ ) = $\frac{\sqrt[]{3} }{2}$ 0 < x < 180 độ .các nghiệm là
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Calantha
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Đáp án:
$x = \left\{30^o;120^0\right\}$
Giải thích các bước giải:
$\begin{array}{l}\sin(x + 30^o) = \dfrac{\sqrt3}{2}\\ \Leftrightarrow \sin(x + 30^o) = \sin60^o\\ \Leftrightarrow \left[\begin{array}{l}x + 30^o = 60^o + k.360^o\\x + 30^o = 120^o + k.360^o\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =30^o + k.360^o\\x =90^o+ k.360^o\end{array}\right.\quad (k \in \Bbb Z)\\Ta\,\,có:\\ 0 < x < 180^o\\ \Leftrightarrow \left[\begin{array}{l}0 <30^o + k.360^o< 180^o\\0<90^o+ k.360^o< 180^o\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}-\dfrac{1}{12}< k < \dfrac{5}{12}\\-\dfrac{1}{4}< k< \dfrac{1}{4}\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}k = 0\\k = 0\end{array}\right.\quad (k \in \Bbb Z)\\ \Rightarrow \left[\begin{array}{l}x = 30^o\\x = 120^o\end{array}\right.\\ Vậy\,\,x = \left\{60^o;120^0\right\} \end{array}$
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