Toán Lớp 11: Giải phương trình sau: $cos(x+120^o)-sin(2x+50^o)=0$
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Toán Lớp 11: Giải phương trình sau: $cos(x+120^o)-sin(2x+50^o)=0$
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\cos \left( {x + {{120}^o}} \right) – \sin \left( {2x + {{50}^o}} \right) = 0\\
\Leftrightarrow \cos \left( {x + {{120}^o}} \right) = \sin \left( {2x + {{50}^o}} \right)\\
\Leftrightarrow \cos \left[ {{{90}^o} – \left( { – x – {{30}^o}} \right)} \right] = \sin \left( {2x + {{50}^o}} \right)\\
\Leftrightarrow \sin \left( { – x – {{30}^o}} \right) = \sin \left( {2x + {{50}^o}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
– x – {30^o} = 2x + {50^o} + k{360^o}\\
– x – {30^o} = {180^o} – 2x – {50^o} + k{360^o}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = – {80^o} – k{360^o}\\
x = {160^o} + k{360^o}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ – {{80}^o}}}{3} – k{120^o}\\
x = {160^o} + k{360^o}
\end{array} \right.\left( {k \in \mathbb{Z}} \right)
\end{array}$