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Khải Quang
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Đáp án:
4) \(\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k2\pi \\
x = – \dfrac{{5\pi }}{{12}} + k2\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\sin 2x = 1\\
\to 2x = \dfrac{\pi }{2} + k2\pi \\
\to x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\
2)\cos \left( {2x + 20^\circ } \right) = – \dfrac{{\sqrt 3 }}{2}\\
\to \left[ \begin{array}{l}
2x + 20^\circ = 150^\circ + k360\\
2x + 20^\circ = – 150^\circ + k360
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{130}}{2} + k180\\
x = – \dfrac{{170}}{2} + k180
\end{array} \right.\left( {k \in Z} \right)\\
3){\sin ^2}x – 3\sin x + 2 = 0\\
\to \left( {\sin x – 1} \right)\left( {\sin x – 2} \right) = 0\\
\to \left[ \begin{array}{l}
\sin x = 1\\
\sin x = 2\left( l \right)
\end{array} \right.\\
\to x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
4)\dfrac{{\sqrt 3 }}{2}.\cos x – \dfrac{1}{2}.\sin x = \dfrac{{\sqrt 2 }}{2}\\
\to \sin \dfrac{\pi }{3}.\cos x – \cos \dfrac{\pi }{3}.\sin x = \dfrac{{\sqrt 2 }}{2}\\
\to \sin \left( {\dfrac{\pi }{3} – x} \right) = \dfrac{{\sqrt 2 }}{2}\\
\to \left[ \begin{array}{l}
\dfrac{\pi }{3} – x = \dfrac{\pi }{4} + k2\pi \\
\dfrac{\pi }{3} – x = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k2\pi \\
x = – \dfrac{{5\pi }}{{12}} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
Đáp án:
Giải thích các bước giải:
Mk k có máy tính bn tự giải tiếp nha