Toán Lớp 11: giải ptr
1 : cos2x – sin2x = 0
2 : 2cos^2x + 5sinx – 4 = 0
3 : 5sinx(sinx-1)-cos^2x=3
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Toán Lớp 11: giải ptr
1 : cos2x – sin2x = 0
2 : 2cos^2x + 5sinx – 4 = 0
3 : 5sinx(sinx-1)-cos^2x=3
Comments ( 1 )
\cos 2x – \sin 2x = 0\\
\Leftrightarrow \sqrt 2 \left( {\dfrac{{\sqrt 2 }}{2}\cos 2x – \dfrac{{\sqrt 2 }}{2}\sin 2x} \right) = 0\\
\Leftrightarrow \sqrt 2 \cos \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = 0\\
\Leftrightarrow x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in \mathbb{Z}} \right)\\
b)2co{s^2}x + 5sinx – 4 = 0\\
\Leftrightarrow 2\left( {{{\cos }^2}x – 1} \right) + 5\sin x – 2 = 0\\
\Leftrightarrow 2\left( { – {{\sin }^2}x} \right) + 5\sin x – 2 = 0\\
\Leftrightarrow 2{\sin ^2} – 5\sin x + 2 = 0\\
\Leftrightarrow \left( {\sin x – 2} \right)\left( {2\sin x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 2(L)\\
\sin x = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \sin x = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
c)5sinx(sinx – 1) – co{s^2}x = 3\\
\Leftrightarrow 5{\sin ^2}x – 5\sin x – {\cos ^2}x – 3 = 0\\
\Leftrightarrow 5{\sin ^2}x – 5\sin x – \left( {{{\cos }^2}x – 1} \right) – 4 = 0\\
\Leftrightarrow 5{\sin ^2}x – 5\sin x + {\sin ^2}x – 4 = 0\\
\Leftrightarrow 6{\sin ^2}x – 5\sin x – 4 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{4}{3}(L)\\
\sin x = \dfrac{{ – 1}}{2}
\end{array} \right.\\
\Leftrightarrow \sin x = – \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\\
\Leftrightarrow x = \dfrac{{ – \pi }}{6} + k\pi \left( {k \in \mathbb{Z}} \right)
\end{array}$