Votee 5 saoo ạ^^ . Mong mng giúp đỡ 2cos^2x + 5sinx – 4= 0 Sin^4x – cos^4x= 2căn3.sinx.cosx + 1 04/09/2021 Bởi Alexandra Votee 5 saoo ạ^^ . Mong mng giúp đỡ 2cos^2x + 5sinx – 4= 0 Sin^4x – cos^4x= 2căn3.sinx.cosx + 1
Đáp án: b. \(\left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{2} + k\pi \end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}a.2{\cos ^2}x + 5\sin x – 4 = 0\\ \to 2\left( {1 – {{\sin }^2}x} \right) + 5\sin x – 4 = 0\\ \to – 2{\sin ^2}x + 5\sin x – 2 = 0\\ \to \left( {2 – \sin x} \right)\left( {2\sin x – 1} \right) = 0\\ \to \left[ \begin{array}{l}\sin x = 2\left( l \right)\\\sin x = \dfrac{1}{2}\end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.\left( {k \in Z} \right)\\b.{\sin ^4}x – {\cos ^4}x = 2\sqrt 3 \sin x.\cos x + 1\\ \to \left( {{{\sin }^2}x – {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = \sqrt 3 \sin 2x + 1\\ \to – \left( {{{\cos }^2}x – {{\sin }^2}x} \right) = \sqrt 3 \sin 2x + 1\\ \to – \cos 2x = \sqrt 3 \sin 2x + 1\\ \to \sqrt 3 \sin 2x + \cos 2x = – 1\\ \to \dfrac{{\sqrt 3 }}{2}\sin 2x + \dfrac{1}{2}\cos 2x = – \dfrac{1}{2}\\ \to \sin 2x.\cos \dfrac{\pi }{6} + \sin \dfrac{\pi }{6}.\cos 2x = – \dfrac{1}{2}\\ \to \sin \left( {2x + \dfrac{\pi }{6}} \right) = – \dfrac{1}{2}\\ \to \left[ \begin{array}{l}2x + \dfrac{\pi }{6} = – \dfrac{\pi }{6} + k2\pi \\2x + \dfrac{\pi }{6} = \dfrac{{7\pi }}{6} + k2\pi \end{array} \right.\\ \to \left[ \begin{array}{l}x = – \dfrac{\pi }{6} + k\pi \\x = \dfrac{\pi }{2} + k\pi \end{array} \right.\left( {k \in Z} \right)\end{array}\) Bình luận
Đáp án:
Giải thích các bước giải:
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Đáp án:
b. \(\left[ \begin{array}{l}
x = – \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.2{\cos ^2}x + 5\sin x – 4 = 0\\
\to 2\left( {1 – {{\sin }^2}x} \right) + 5\sin x – 4 = 0\\
\to – 2{\sin ^2}x + 5\sin x – 2 = 0\\
\to \left( {2 – \sin x} \right)\left( {2\sin x – 1} \right) = 0\\
\to \left[ \begin{array}{l}
\sin x = 2\left( l \right)\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
b.{\sin ^4}x – {\cos ^4}x = 2\sqrt 3 \sin x.\cos x + 1\\
\to \left( {{{\sin }^2}x – {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = \sqrt 3 \sin 2x + 1\\
\to – \left( {{{\cos }^2}x – {{\sin }^2}x} \right) = \sqrt 3 \sin 2x + 1\\
\to – \cos 2x = \sqrt 3 \sin 2x + 1\\
\to \sqrt 3 \sin 2x + \cos 2x = – 1\\
\to \dfrac{{\sqrt 3 }}{2}\sin 2x + \dfrac{1}{2}\cos 2x = – \dfrac{1}{2}\\
\to \sin 2x.\cos \dfrac{\pi }{6} + \sin \dfrac{\pi }{6}.\cos 2x = – \dfrac{1}{2}\\
\to \sin \left( {2x + \dfrac{\pi }{6}} \right) = – \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
2x + \dfrac{\pi }{6} = – \dfrac{\pi }{6} + k2\pi \\
2x + \dfrac{\pi }{6} = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{\pi }{6} + k\pi \\
x = \dfrac{\pi }{2} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)