2 câu trả lời
Đáp án:
$\begin{array}{l}
\cos \left( {2x + {{60}^0}} \right) = \cos x\\
\Rightarrow \left[ \begin{array}{l}
2x + {60^0} = x + k{.360^0}\\
2x + {60^0} = - x + k{.360^0}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - {60^0} + k{.360^0}\\
x = - {20^0} + k{.120^0}
\end{array} \right.
\end{array}$
`cos (2x+60^o)=cos x`
`⇔ ` \(\left[ \begin{array}{l}2x+60^o=x+k.360^o\\2x+60^o=-x+k.360^o\end{array} \right.\)
`⇔ ` \(\left[ \begin{array}{l}2x-x=-60^o+k.360^o\\2x+x=-60^o+k.360^o\end{array} \right.\)
`⇔ ` \(\left[ \begin{array}{l}x=-60^o+k.360^o\\3x=-60^o+k.360^o\end{array} \right.\)
`⇔ ` \(\left[ \begin{array}{l}x=-60^o+k.360^o\\x=-20^o+k.120^o\end{array} \right.\) $(k∈Z)$
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