Toán 1:sin x=cos 2x 2: sin(3x+25°)=-√3/2 3: sin x = cos 1/x 13/07/2021 By Sadie 1:sin x=cos 2x 2: sin(3x+25°)=-√3/2 3: sin x = cos 1/x
Đáp án: 2. \(\left[ \begin{array}{l}x=-\dfrac{17\pi}{108}+\dfrac{k2\pi}{3}\\x=\dfrac{49\pi}{108}+\dfrac{k2\pi}{3}\end{array} \right.\) Giải thích các bước giải: 1. $sin x= 1-2sinx^2$ ⇔ $2sinx^2+ sin x-1=0⇔$ \(\left[ \begin{array}{l}sin x=\dfrac{1}{2}\\sin x=-1\end{array} \right.\) Với $sin x= \dfrac{1}{2}⇒$ \(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+ k2\pi\\x=\dfrac{5\pi}{6}+ k2\pi\end{array} \right.\) Với $sin x= -1$ ⇒\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+ k2\pi\\x=\dfrac{3\pi}{2}+ k2\pi\end{array} \right.\) 2. $sin ( 3x+ \dfrac{5\pi}{36})=sin (-\dfrac{\pi}{3})$ ⇔\(\left[ \begin{array}{l}3x+\dfrac{5\pi}{36}=-\dfrac{\pi}{3}+ k2\pi\\3x+\dfrac{5\pi}{36}=\dfrac{4\pi}{3}+k2\pi\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=-\dfrac{17\pi}{108}+\dfrac{k2\pi}{3}\\x=\dfrac{49\pi}{108}+\dfrac{k2\pi}{3}\end{array} \right.\) 3. $sin x= sin (\dfrac{\pi}{2}-\dfrac{1}{x})$ \(\left[ \begin{array}{l}x= \dfrac{\pi}{2}-\dfrac{1}{x}+k2\pi\\x=\dfrac{\pi}{2}+\dfrac{1}{x}+ k2\pi\end{array} \right.\) Trả lời
`~rai~` \(1.sinx=cos2x\\\Leftrightarrow sinx=sin\left(\dfrac{\pi}{2}-2x\right)\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}-2x+k2\pi\\x=\pi-\dfrac{\pi}{2}+2x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}3x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{2}-k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{6}+k\dfrac{2\pi}{3}\\x=-\dfrac{\pi}{2}-k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\2.sin\left(3x+25^0\right)=-\dfrac{\sqrt{3}}{2}\\\Leftrightarrow sin\left(3x+25^0\right)=sin(-60^0)\\\Leftrightarrow \left[\begin{array}{I}3x+25^0=-60^0+k360^0\\3x+25^0=180^0-(-60^0)+k360^0\end{array}\right.\\\Leftrightarrow \left[\begin{array}3x=-85^0+k360^0\\3x=215^0+k360^0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{85}{3}^0+k120^0\\x=\dfrac{215}{3}^0+k120^0.\end{array}\right.\quad(k\in\mathbb{Z}) \) Trả lời
Đáp án:
2. \(\left[ \begin{array}{l}x=-\dfrac{17\pi}{108}+\dfrac{k2\pi}{3}\\x=\dfrac{49\pi}{108}+\dfrac{k2\pi}{3}\end{array} \right.\)
Giải thích các bước giải:
1. $sin x= 1-2sinx^2$
⇔ $2sinx^2+ sin x-1=0⇔$ \(\left[ \begin{array}{l}sin x=\dfrac{1}{2}\\sin x=-1\end{array} \right.\)
Với $sin x= \dfrac{1}{2}⇒$ \(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+ k2\pi\\x=\dfrac{5\pi}{6}+ k2\pi\end{array} \right.\)
Với $sin x= -1$ ⇒\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+ k2\pi\\x=\dfrac{3\pi}{2}+ k2\pi\end{array} \right.\)
2. $sin ( 3x+ \dfrac{5\pi}{36})=sin (-\dfrac{\pi}{3})$
⇔\(\left[ \begin{array}{l}3x+\dfrac{5\pi}{36}=-\dfrac{\pi}{3}+ k2\pi\\3x+\dfrac{5\pi}{36}=\dfrac{4\pi}{3}+k2\pi\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-\dfrac{17\pi}{108}+\dfrac{k2\pi}{3}\\x=\dfrac{49\pi}{108}+\dfrac{k2\pi}{3}\end{array} \right.\)
3. $sin x= sin (\dfrac{\pi}{2}-\dfrac{1}{x})$
\(\left[ \begin{array}{l}x= \dfrac{\pi}{2}-\dfrac{1}{x}+k2\pi\\x=\dfrac{\pi}{2}+\dfrac{1}{x}+ k2\pi\end{array} \right.\)
`~rai~`
\(1.sinx=cos2x\\\Leftrightarrow sinx=sin\left(\dfrac{\pi}{2}-2x\right)\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{2}-2x+k2\pi\\x=\pi-\dfrac{\pi}{2}+2x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}3x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{2}-k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{6}+k\dfrac{2\pi}{3}\\x=-\dfrac{\pi}{2}-k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\2.sin\left(3x+25^0\right)=-\dfrac{\sqrt{3}}{2}\\\Leftrightarrow sin\left(3x+25^0\right)=sin(-60^0)\\\Leftrightarrow \left[\begin{array}{I}3x+25^0=-60^0+k360^0\\3x+25^0=180^0-(-60^0)+k360^0\end{array}\right.\\\Leftrightarrow \left[\begin{array}3x=-85^0+k360^0\\3x=215^0+k360^0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{85}{3}^0+k120^0\\x=\dfrac{215}{3}^0+k120^0.\end{array}\right.\quad(k\in\mathbb{Z}) \)