Mn giúp e lm mấy ý này với ạ a) sin (2x-1)= sin (3x+1) b) sin (8x +60°)+sin 2x=0 c) sin 5x = -sin 2x d) sin 3x = sin x

Đáp án:

 $a)  {\left[\begin{aligned}x=-4-k2\pi\\x=\dfrac{\pi+k2\pi}{5}\end{aligned}\right.} (k\in \mathbb{Z})\\
b) 
 {\left[\begin{aligned} x=-6^{\circ}+k.36^{\circ}\\x=20^{\circ}+k.60^{\circ}\end{aligned}\right.},(k\in \mathbb{Z})\\
c) 
 {\left[\begin{aligned} x=\dfrac{k2\pi}{7}\\ x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\end{aligned}\right.},(k\in \mathbb{Z})\\
d)
{\left[\begin{aligned} x=k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{aligned}\right.}(k\in\mathbb{Z})\\$

Giải thích các bước giải:

 $a) \sin(2x-1)=\sin(3x+1)\\
\Leftrightarrow {\left[\begin{aligned}2x-1=3x+1+k2\pi\\2x-1=\pi-3x-1+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x-3x=1+1+k2\pi\\2x+3x=\pi+1-1+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}-x=4+k2\pi\\5x=\pi+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=-4-k2\pi\\x=\dfrac{\pi+k2\pi}{5}\end{aligned}\right.} (k\in \mathbb{Z})\\
b) 
\sin(8x+60^{\circ})+\sin2x=0\\
\Leftrightarrow \sin(8x+60^{\circ})=-\sin2x\\
\Leftrightarrow \sin(8x+60^{\circ})=\sin(-2x)\\
\Leftrightarrow {\left[\begin{aligned}8x+60^{\circ}=-2x+k.360^{\circ}\\8x+60^{\circ}=180^{\circ}+2x+k.360^{\circ}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}8x+2x=-60^{\circ}+k.360^{\circ}\\8x-2x=180^{\circ}-60^{\circ}+k.360^{\circ}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 10x=-60^{\circ}+k.360^{\circ}\\6x=120^{\circ}+k.360^{\circ}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=-6^{\circ}+k.36^{\circ}\\x=20^{\circ}+k.60^{\circ}\end{aligned}\right.},(k\in \mathbb{Z})\\
c) 
\sin5x=-\sin2x\\
\Leftrightarrow \sin5x=\sin(-2x)\\
\Leftrightarrow {\left[\begin{aligned} 5x=-2x+k2\pi\\ 5x=\pi+2x+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 5x+2x=k2\pi\\ 5x-2x=\pi+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 7x=k2\pi\\ 3x=\pi+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=\dfrac{k2\pi}{7}\\ x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\end{aligned}\right.},(k\in \mathbb{Z})\\
d)
\sin3x=\sin x\\
\Leftrightarrow {\left[\begin{aligned} 3x=x+k2\pi\\3x=\pi-x+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 3x-x=k2\pi\\3x+x=\pi+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} 2x=k2\pi\\4x=\pi+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned} x=k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{aligned}\right.}(k\in\mathbb{Z})\\$