a) Cosx -sinx= căn 2 b) 2sin^2x -sinxcosx – cos^2x =2 c) căn3 sin^2x – sinxcosx=0 d) 1- căn2 × sin × (4x -20°) =0 e) y= 2tan (pi/7 – 2x)+1 (Tìm TXĐ hà

0 bình luận về “a) Cosx -sinx= căn 2 b) 2sin^2x -sinxcosx – cos^2x =2 c) căn3 sin^2x – sinxcosx=0 d) 1- căn2 × sin × (4x -20°) =0 e) y= 2tan (pi/7 – 2x)+1 (Tìm TXĐ hà”

1. \(\begin{array}{l} a)\cos x – \sin x = \sqrt 2 \\ \Leftrightarrow \sqrt 2 \cos \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 \Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = k2\pi \Leftrightarrow x = – \dfrac{\pi }{4} + k2\pi \\ b)2{\sin ^2}x – \sin x\cos x – {\cos ^2}x = 2\\ \Leftrightarrow 1 – \cos 2x – \dfrac{1}{2}\sin 2x – \dfrac{{1 + \cos 2x}}{2} = 2\\ \Leftrightarrow 2 – 2\cos 2x – \sin 2x – 1 – \cos 2x – 4 = 0\\ \Leftrightarrow – \sin 2x – 3\cos 2x – 3 = 0\\ \Leftrightarrow \sin 2x + 3\cos 2x = – 3\\ \Leftrightarrow \dfrac{1}{{\sqrt {10} }}\sin 2x + \dfrac{3}{{\sqrt {10} }}\cos 2x = – \dfrac{3}{{\sqrt {10} }}\\ \Leftrightarrow \sin \left( {2x + \alpha } \right) = – \sin \alpha = \sin \left( { – \alpha } \right)\\ \Leftrightarrow \left[ \begin{array}{l} 2x + \alpha = – \alpha + k2\pi \\ 2x + \alpha = \pi + \alpha + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = – \alpha + k\pi \\ x = \dfrac{\pi }{2} + k\pi \end{array} \right. \end{array}\)

   Vậy $x = – \alpha + k\pi $ và $ x = \dfrac{\pi }{2} + k\pi$ $(k\in\mathbb Z)$.

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2. c) Rút $\sin x$ ra ngoài ta có

   $\sin x(\sqrt{3} \sin x – \cos x) = 0$

   Vậy $\sin x = 0$ hay $x = k\pi$ hoặc

   $\sqrt{3} \sin x – \cos x = 0$

   $\Leftrightarrow \sin x \dfrac{\sqrt{3}}{2} – \dfrac{1}{2} \cos x = 0$

   $\Leftrightarrow \sin (x – \dfrac{\pi}{6}) = 0$

   Vậy $x – \dfrac{\pi}{6} = k\pi$ hay $x = \dfrac{pi}{6} + k\pi$.

   Vậy $x = k\pi$ và $x = \dfrac{\pi}{6} + k\pi$.

   d) $1-\sqrt2\sin(4x-20^o)=0$

   $\Leftrightarrow \sin(4x-\dfrac{\pi}9)=\dfrac1{\sqrt2}$

   $\Leftrightarrow 4x-\dfrac{\pi}9=\dfrac{\pi}4+k2\pi$ hoặc $4x-\dfrac{\pi}9=\pi-\dfrac{\pi}4+k2\pi$

   $\Leftrightarrow x=\dfrac{13\pi}{144}+k\dfrac{\pi}2$ hoặc $x=\dfrac{31\pi}{144}$ $(k\in\mathbb Z)$

   Vậy $x=\dfrac{13\pi}{144}+k\dfrac{\pi}2$ và $x=\dfrac{31\pi}{144}$ $(k\in\mathbb Z)$

   e) Điều kiện $\cos(\dfrac{\pi}{7} – 2x) \neq 0$

   Vậy $\dfrac{\pi}{7} – 2x \neq \dfrac{\pi}{2} + k\pi$

   Do đó $x \neq -\dfrac{5\pi}{14} + k\pi$.

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