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 hàm số)
2 câu trả lời
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 $x \neq -\dfrac{5\pi}{14} + k\pi$.
\(\begin{array}{l} a)\cos x - \sin x = \sqrt 2 \\ \Leftrightarrow \sqrt 2 \cos \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 \Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = k2\pi \Leftrightarrow x = - \dfrac{\pi }{4} + k2\pi \\ b)2{\sin ^2}x - \sin x\cos x - {\cos ^2}x = 2\\ \Leftrightarrow 1 - \cos 2x - \dfrac{1}{2}\sin 2x - \dfrac{{1 + \cos 2x}}{2} = 2\\ \Leftrightarrow 2 - 2\cos 2x - \sin 2x - 1 - \cos 2x - 4 = 0\\ \Leftrightarrow - \sin 2x - 3\cos 2x - 3 = 0\\ \Leftrightarrow \sin 2x + 3\cos 2x = - 3\\ \Leftrightarrow \dfrac{1}{{\sqrt {10} }}\sin 2x + \dfrac{3}{{\sqrt {10} }}\cos 2x = - \dfrac{3}{{\sqrt {10} }}\\ \Leftrightarrow \sin \left( {2x + \alpha } \right) = - \sin \alpha = \sin \left( { - \alpha } \right)\\ \Leftrightarrow \left[ \begin{array}{l} 2x + \alpha = - \alpha + k2\pi \\ 2x + \alpha = \pi + \alpha + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - \alpha + k\pi \\ x = \dfrac{\pi }{2} + k\pi \end{array} \right. \end{array}\)
Vậy $x = - \alpha + k\pi $ và $ x = \dfrac{\pi }{2} + k\pi$ $(k\in\mathbb Z)$.
