Đáp án: \[x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\,\,\,\,\,\,\,\left( {k \in Z} \right)\] Giải thích các bước giải: Ta có: \(\begin{array}{l}{\sin ^2}x + {\sin ^2}2x = 1\\ \Leftrightarrow {\sin ^2}x + {\left( {2\sin x.\cos x} \right)^2} = 1\\ \Leftrightarrow {\sin ^2}x + 4{\sin ^2}x.{\cos ^2}x = 1\\ \Leftrightarrow {\sin ^2}x + 4.{\sin ^2}x.\left( {1 – {{\sin }^2}x} \right) = 1\\ \Leftrightarrow {\sin ^2}x + 4{\sin ^2}x – 4{\sin ^4}x = 1\\ \Leftrightarrow 4{\sin ^4}x – 5{\sin ^2}x + 1 = 0\\ \Leftrightarrow \left( {{{\sin }^2}x – 1} \right)\left( {4{{\sin }^2}x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}{\sin ^2}x = 1\\{\sin ^2}x = \dfrac{1}{4}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}{\cos ^2}x = 0\\\sin x = \dfrac{1}{2}\\\sin x = – \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = – \dfrac{\pi }{2} + k\pi \\x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \\x = – \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\,\,\,\,\,\,\,\left( {k \in Z} \right)\end{array}\) Reply
Đáp án:
\[x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\,\,\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x + {\sin ^2}2x = 1\\
\Leftrightarrow {\sin ^2}x + {\left( {2\sin x.\cos x} \right)^2} = 1\\
\Leftrightarrow {\sin ^2}x + 4{\sin ^2}x.{\cos ^2}x = 1\\
\Leftrightarrow {\sin ^2}x + 4.{\sin ^2}x.\left( {1 – {{\sin }^2}x} \right) = 1\\
\Leftrightarrow {\sin ^2}x + 4{\sin ^2}x – 4{\sin ^4}x = 1\\
\Leftrightarrow 4{\sin ^4}x – 5{\sin ^2}x + 1 = 0\\
\Leftrightarrow \left( {{{\sin }^2}x – 1} \right)\left( {4{{\sin }^2}x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\sin ^2}x = 1\\
{\sin ^2}x = \dfrac{1}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{\cos ^2}x = 0\\
\sin x = \dfrac{1}{2}\\
\sin x = – \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi \\
x = – \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right. \Leftrightarrow x = \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\,\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)