Toán a) 2sin(x-π/6)+1=0 b) √3cos2x+sin2x-1=0 @pu vi 25/11/2021 By Brielle a) 2sin(x-π/6)+1=0 b) √3cos2x+sin2x-1=0 @pu vi
Đáp án: $a) 2sin(x-π/6)+1=0$ $⇔sin(x-\dfrac{\pi}{6})=$$\dfrac{-1}{2}$ $⇔$\(\left[ \begin{array}{l}x-\dfrac{\pi}{6}=\dfrac{-\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{7\pi}{6}+k2\pi\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x=k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{array} \right.\) $(k∈Z)$ $b) √3cos2x+sin2x-1=0$ Chia cả hai vế cho $\sqrt[]{a^2+b^2}=2$ $⇒\dfrac{√3}{2}cos2x+$$\dfrac{1}{2}sin2x=$$\dfrac{1}{2}$ $⇔sin(\dfrac{\pi}{3}).cos2x+sin2x.cos\dfrac{\pi}{3}=\dfrac{1}{2}$ $⇔sin(\dfrac{\pi}{3}+2x)=\dfrac{1}{2}$ $⇔$\(\left[ \begin{array}{l}\dfrac{\pi}{3}+2x=\dfrac{\pi}{6}+k2\pi\\\dfrac{\pi}{3}+2x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\) $⇔$\(\left[ \begin{array}{l}x=\dfrac{-\pi}{12}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{array} \right.\) $(k ∈Z)$ BẠN THAM KHẢO NHA!!! Trả lời
Đáp án: a) $\left[\begin{array}{l}x = k2\pi\\x= \dfrac{4\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$ b) $\left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{12}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$ Giải thích các bước giải: a) $2\sin\left(x -\dfrac{\pi}{6}\right) + 1 = 0$ $\to \sin\left(x -\dfrac{\pi}{6}\right) = -\dfrac12$ $\to \sin\left(x -\dfrac{\pi}{6}\right) = \sin\left(-\dfrac{\pi}{6}\right)$ $\to \left[\begin{array}{l}x -\dfrac{\pi}{6} = -\dfrac{\pi}{6} + k2\pi\\x -\dfrac{\pi}{6} = \dfrac{7\pi}{6} + k2\pi\end{array}\right.$ $\to \left[\begin{array}{l}x = k2\pi\\x= \dfrac{4\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$ b) $\sqrt3\cos2x + \sin2x -1 = 0$ $\to \dfrac{\sqrt3}{2}\cos2x +\dfrac12\sin2x =\dfrac12$ $\to \cos\left(2x -\dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{3}$ $\to \left[\begin{array}{l}2x -\dfrac{\pi}{6}=\dfrac{\pi}{3} + k2\pi\\2x -\dfrac{\pi}{6} = -\dfrac{\pi}{3}+ k2\pi\end{array}\right.$ $\to \left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{12}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$ Trả lời
Đáp án:
$a) 2sin(x-π/6)+1=0$
$⇔sin(x-\dfrac{\pi}{6})=$$\dfrac{-1}{2}$
$⇔$\(\left[ \begin{array}{l}x-\dfrac{\pi}{6}=\dfrac{-\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{7\pi}{6}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{array} \right.\) $(k∈Z)$
$b) √3cos2x+sin2x-1=0$ Chia cả hai vế cho $\sqrt[]{a^2+b^2}=2$
$⇒\dfrac{√3}{2}cos2x+$$\dfrac{1}{2}sin2x=$$\dfrac{1}{2}$
$⇔sin(\dfrac{\pi}{3}).cos2x+sin2x.cos\dfrac{\pi}{3}=\dfrac{1}{2}$
$⇔sin(\dfrac{\pi}{3}+2x)=\dfrac{1}{2}$
$⇔$\(\left[ \begin{array}{l}\dfrac{\pi}{3}+2x=\dfrac{\pi}{6}+k2\pi\\\dfrac{\pi}{3}+2x=\dfrac{5\pi}{6}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{-\pi}{12}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{array} \right.\) $(k ∈Z)$
BẠN THAM KHẢO NHA!!!
Đáp án:
a) $\left[\begin{array}{l}x = k2\pi\\x= \dfrac{4\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{12}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
a) $2\sin\left(x -\dfrac{\pi}{6}\right) + 1 = 0$
$\to \sin\left(x -\dfrac{\pi}{6}\right) = -\dfrac12$
$\to \sin\left(x -\dfrac{\pi}{6}\right) = \sin\left(-\dfrac{\pi}{6}\right)$
$\to \left[\begin{array}{l}x -\dfrac{\pi}{6} = -\dfrac{\pi}{6} + k2\pi\\x -\dfrac{\pi}{6} = \dfrac{7\pi}{6} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = k2\pi\\x= \dfrac{4\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\sqrt3\cos2x + \sin2x -1 = 0$
$\to \dfrac{\sqrt3}{2}\cos2x +\dfrac12\sin2x =\dfrac12$
$\to \cos\left(2x -\dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{3}$
$\to \left[\begin{array}{l}2x -\dfrac{\pi}{6}=\dfrac{\pi}{3} + k2\pi\\2x -\dfrac{\pi}{6} = -\dfrac{\pi}{3}+ k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{12}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$