Toán Lớp 11: 1, sin^2x-sin2x+cos^2x-1=0
2, cos^2x-2sinxcosx-sin^2x=0
3, sin^2x-3sinxcosx=1
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Toán Lớp 11: 1, sin^2x-sin2x+cos^2x-1=0
2, cos^2x-2sinxcosx-sin^2x=0
3, sin^2x-3sinxcosx=1
Comments ( 2 )
1)\quad S = \left\{k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
2)\quad S = \left\{\dfrac{\pi}{8} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
3)\quad S = \left\{\dfrac{\pi}{2} + k\pi;\ \arctan\left(-\dfrac13\right) + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\end{array}\)
1)\\
\quad \sin^2x -\sin2x + \cos^2x – 1 =0\\
\Leftrightarrow \sin2x = 0\\
\Leftrightarrow 2x = k\pi\\
\Leftrightarrow x = k\dfrac{\pi}{2}\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
2)\\
\quad \cos^2x – 2\sin x\cos x -\sin^2x = 0\\
\Leftrightarrow \cos2x – \sin2x = 0\\
\Leftrightarrow \sqrt\cos\left(2x + \dfrac{\pi}{4}\right)=0\\
\Leftrightarrow 2x + \dfrac{\pi}{4} = \dfrac{\pi}{2} + k\pi\\
\Leftrightarrow x = \dfrac{\pi}{8} + k\dfrac{\pi}{2}\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{8} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}
3)\\
\quad \sin^2x -3\sin x\cos x = 1\qquad (2)\\
\bullet\quad \cos x = 0\Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k\in\Bbb Z)\\
(*) \Leftrightarrow \sin^2x = 1\\
\Leftrightarrow \sin x = \pm 1\\
\Leftrightarrow x = \dfrac{\pi}{2} + k\pi\quad (k\in\Bbb Z)\\
\Rightarrow x = \dfrac{\pi}{2} + k\pi\ \text{là một họ nghiệm của $(1)$}\\
\bullet\quad \cos x \ne 0\\
\text{Chia hai vế của phương trình cho $\cos^2x$ ta được:}\\
\quad \tan^2x – 3\tan x = \dfrac{1}{\cos^2x}\\
\Leftrightarrow \tan^2x – 3\tan x = \tan^2 x+ 1\\
\Leftrightarrow \tan x = -\dfrac13\\
\Leftrightarrow x = \arctan\left(-\dfrac13\right) + k\pi\quad (k\in\Bbb Z)\\
\text{Vậy}\ S = \left\{\dfrac{\pi}{2} + k\pi;\ \arctan\left(-\dfrac13\right) + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\end{array}\)