Toán 1) sin2x-2cos^2 2x-1=0 2) căn3 cos7x-2sin5xcos2x-sin3x=0 10/09/2021 By Kinsley 1) sin2x-2cos^2 2x-1=0 2) căn3 cos7x-2sin5xcos2x-sin3x=0
Đáp án: \(\begin{array}{l} 1)\,\,\,x = \frac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right).\\ 2)\,\,x = \frac{\pi }{{21}} + \frac{{k\pi }}{7}\,\,\,\left( {k \in Z} \right) \end{array}\) Giải thích các bước giải: \(\begin{array}{l} 1)\,\,\sin 2x – 2{\cos ^2}2x – 1 = 0\\ \Leftrightarrow \sin 2x – 2\left( {1 – {{\sin }^2}2x} \right) – 1 = 0\\ \Leftrightarrow 2{\sin ^2}2x + \sin 2x – 3 = 0\\ \Leftrightarrow \left( {2\sin 2x + 3} \right)\left( {\sin 2x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin 2x = 1\\ \sin 2x = – \frac{3}{2}\,\,\,\left( {ktm} \right) \end{array} \right. \Leftrightarrow 2x = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right).\\ 2)\,\,\sqrt 3 \cos 7x – 2\sin 5x.\cos 2x – \sin 3x = 0\\ \Leftrightarrow \,\sqrt 3 \cos 7x – \left( {\sin 3x + \sin 7x} \right) – \sin 3x = 0\\ \Leftrightarrow \sqrt 3 \cos 7x – \sin 3x – \sin 7x – \sin 3x = 0\\ \Leftrightarrow \sqrt 3 \cos 7x – \sin 7x = 0\\ \Leftrightarrow \frac{{\sqrt 3 }}{2}\cos 7x – \frac{1}{2}\sin 7x = 0\\ \Leftrightarrow \cos \left( {7x + \frac{\pi }{6}} \right) = 0\\ \Leftrightarrow 7x + \frac{\pi }{6} = \frac{\pi }{2} + k\pi \\ \Leftrightarrow x = \frac{\pi }{{21}} + \frac{{k\pi }}{7}\,\,\,\left( {k \in Z} \right). \end{array}\) Trả lời
Đáp án:
\(\begin{array}{l}
1)\,\,\,x = \frac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right).\\
2)\,\,x = \frac{\pi }{{21}} + \frac{{k\pi }}{7}\,\,\,\left( {k \in Z} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\,\,\sin 2x – 2{\cos ^2}2x – 1 = 0\\
\Leftrightarrow \sin 2x – 2\left( {1 – {{\sin }^2}2x} \right) – 1 = 0\\
\Leftrightarrow 2{\sin ^2}2x + \sin 2x – 3 = 0\\
\Leftrightarrow \left( {2\sin 2x + 3} \right)\left( {\sin 2x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 1\\
\sin 2x = – \frac{3}{2}\,\,\,\left( {ktm} \right)
\end{array} \right. \Leftrightarrow 2x = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{\pi }{4} + k\pi \,\,\,\left( {k \in Z} \right).\\
2)\,\,\sqrt 3 \cos 7x – 2\sin 5x.\cos 2x – \sin 3x = 0\\
\Leftrightarrow \,\sqrt 3 \cos 7x – \left( {\sin 3x + \sin 7x} \right) – \sin 3x = 0\\
\Leftrightarrow \sqrt 3 \cos 7x – \sin 3x – \sin 7x – \sin 3x = 0\\
\Leftrightarrow \sqrt 3 \cos 7x – \sin 7x = 0\\
\Leftrightarrow \frac{{\sqrt 3 }}{2}\cos 7x – \frac{1}{2}\sin 7x = 0\\
\Leftrightarrow \cos \left( {7x + \frac{\pi }{6}} \right) = 0\\
\Leftrightarrow 7x + \frac{\pi }{6} = \frac{\pi }{2} + k\pi \\
\Leftrightarrow x = \frac{\pi }{{21}} + \frac{{k\pi }}{7}\,\,\,\left( {k \in Z} \right).
\end{array}\)