Toán Lớp 11: Giải pt:
a)2cos2x-1-sin2x=2(sinx+cosx)
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Toán Lớp 11: Giải pt:
a)2cos2x-1-sin2x=2(sinx+cosx)
Comments ( 1 )
\quad 2\cos2x – 1 – \sin2x = 2(\sin x + \cos x)\\
\Leftrightarrow 2(\cos^2x – \sin^2x) – (1 +2\sin x\cos x) – 2(\sin x + \cos x) = 0\\
\Leftrightarrow 2(\cos x – \sin x)(\cos x + \sin x) – (\sin x + \cos x)^2 – 2(\sin x + \cos x) =0\\
\Leftrightarrow (\sin x + \cos x)[2(\cos x – \sin x)- (\sin+\cos x) – 2] = 0\\
\Leftrightarrow \sqrt2\sin\left(x + \dfrac{\pi}{4}\right)(\cos x – 3\sin x – 2) =0\\
\Leftrightarrow \left[\begin{array}{l}\sin\left(x + \dfrac{\pi}{4}\right) = 0\\\cos x – 3\sin x = 2\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x + \dfrac{\pi}{4} = k\pi\\\dfrac{1}{\sqrt{10}}\cos x – \dfrac{3}{\sqrt{10}}\sin x = \dfrac{2}{\sqrt{10}}\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{4} + k\pi\\\cos(x + \alpha) = \dfrac{2}{\sqrt{10}}\end{array}\right.\ \ \text{với}\ \begin{cases}\cos\alpha = \dfrac{1}{\sqrt{10}}\\\sin\alpha = \dfrac{3}{\sqrt{10}}\end{cases}\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{4} + k\pi\\x + \alpha = \arccos\dfrac{2}{\sqrt{10}} + k2\pi\\x + \alpha = -\arccos\dfrac{2}{\sqrt{10}} + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = – \dfrac{\pi}{4} + k\pi\\x = \arccos\dfrac{2}{\sqrt{10}}-\alpha + k2\pi\\x = -\arccos\dfrac{2}{\sqrt{10}}-\alpha + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có các họ nghiệm là}\\
x=- \dfrac{\pi}{4} + k\pi\\x = \arccos\dfrac{2}{\sqrt{10}}-\alpha + k2\pi\\x = -\arccos\dfrac{2}{\sqrt{10}}-\alpha + k2\pi\\
\text{với $\alpha$ thỏa mãn}\ \begin{cases}\cos\alpha = \dfrac{1}{\sqrt{10}}\\\sin\alpha = \dfrac{3}{\sqrt{10}}\end{cases}\ \text{và}\ k\in\Bbb Z
\end{array}\)