Toán Lớp 11: 1. Giải pt sau
(1-√2)(1+sinx+cosx)=sin2x
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Toán Lớp 11: 1. Giải pt sau
(1-√2)(1+sinx+cosx)=sin2x
Comments ( 2 )
(sinx -cosx )=t
sin2x =1-t^2
<=>(1-√2)(1+t)=1-t^2
<=t^2-(√2-1)t -√2=0
(t+1)(t-√2)=0
t=-1=>x=3π/2+k2π; x=k2π{k€z)
t=√2
x=3π/4+kπ
\quad \left(1 – \sqrt2\right)\left(1 + \sin x + \cos x\right) = \sin2x\\
\text{Đặt}\ t = \sin x + \cos x\quad \left(|t| \leqslant \sqrt2\right)\\
\Rightarrow t^2 = 1 + \sin2x\\
\Rightarrow t^2 – 1 = \sin2x\\
\text{Phương trình trở thành:}\\
\quad \left(1 – \sqrt2\right)(1 + t) = t^2 – 1\\
\Leftrightarrow t^2 – \left(1-\sqrt2\right)t – 2 + \sqrt2 =0\\
\Leftrightarrow \left[\begin{array}{l}t = -1\\t = 2- \sqrt2\end{array}\right.\\
+)\quad \text{Với $t = -1$ ta được:}\\
\quad \sin x + \cos x = -1\\
\Leftrightarrow \sqrt2\sin\left(x + \dfrac{\pi}{4}\right) = -1\\
\Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right) = -\dfrac{\sqrt2}{2}\\
\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = -\dfrac{\pi}{4} + k2\pi\\x + \dfrac{\pi}{4} = \dfrac{5\pi}{4} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{2} + k2\pi\\x = \pi + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
+)\quad \text{Với $t = 2 -\sqrt2$ ta được:}\\
\quad \sin x + \cos x =2 -\sqrt2\\
\Leftrightarrow \sqrt2\sin\left(x + \dfrac{\pi}{4}\right) = 2-\sqrt2\\
\Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right) =\sqrt2 -1\\
\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = \arcsin\left(\sqrt2 – 1\right) + k2\pi\\x + \dfrac{\pi}{4} = \pi – \arcsin\left(\sqrt2 – 1\right)+ k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x =\arcsin\left(\sqrt2 – 1\right) – \dfrac{\pi}{4} + k2\pi\\x =\dfrac{3\pi}{4}-\arcsin\left(\sqrt2 – 1\right) + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\end{array}\)