Toán Lớp 11: Giải pt lượng giác sau;
Sinx + cosx =-1
Sin^2x -4sinx +3 =0
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Toán Lớp 11: Giải pt lượng giác sau;
Sinx + cosx =-1
Sin^2x -4sinx +3 =0
Comments ( 1 )
a)\sin x + \cos x = – 1\\
\Leftrightarrow \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = – 1\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = – \dfrac{1}{{\sqrt 2 }}\\
\Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin \left( { – \dfrac{\pi }{4}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = – \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{4} = \pi + \dfrac{\pi }{4} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x = – \dfrac{\pi }{2} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
Vậy\,x = – \dfrac{\pi }{2} + k2\pi ;x = \pi + k2\pi \left( {k \in Z} \right)\\
b){\sin ^2}x – 4\sin x + 3 = 0\\
\Leftrightarrow \left( {\sin x – 3} \right)\left( {\sin x – 1} \right) = 0\\
\Leftrightarrow \sin x = 1\left( {do: – 1 \le \sin x \le 1} \right)\\
\Leftrightarrow x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\\
Vậy\,x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)
\end{array}$