Bài 2: Phương trình lượng giác cơ bản

\(sin3x+sin\left(5x-\dfrac{\pi}{6}\right)=0.\\ TXD:D=R.\\ \Leftrightarrow sin3x=-sin\left(5x-\dfrac{\pi}{6}\right).\\ \Leftrightarrow sin3x=sin\left(\dfrac{\pi}{6}+5x\right).\\ \Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+5x+k2\pi.\\3x=\pi-\dfrac{\pi}{6}-5x+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=\dfrac{\pi}{6}+k2\pi\\8x=\dfrac{5}{6}\pi+k2\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{12}\pi-k\pi.\\x=\dfrac{5}{48}\pi+\dfrac{k\pi}{4}.\end{matrix}\right.\)

Lời giải:
\(\sin 3x+\sin (5x-\frac{\pi}{6})=0\)

\(\Leftrightarrow \sin (5x-\frac{\pi}{6})=-\sin 3x=\sin (-3x)\)

\(\Leftrightarrow 5x-\frac{\pi}{6}=-3x+2k\pi\) hoặc $5x-\frac{\pi}{6}=\pi +3x+2k\pi$ với $k$ nguyên

$\Leftrightarrow x=\frac{1}{8}(2k+\frac{1}{6})\pi$ hoặc $x=\frac{1}{2}(\frac{7}{6}+2k)\pi$ với $k$ nguyên.

\(sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=\dfrac{-1}{2}.\\ TXD:D=R.\\ \Leftrightarrow sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=sin\dfrac{-\pi}{6}.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{\pi}{4}=\dfrac{-\pi}{6}+k2\pi.\\\dfrac{x}{2}-\dfrac{\pi}{4}=\pi-\dfrac{-\pi}{6}+k2\pi.\end{matrix}\right.\) \(\left(k\in Z\right).\)

\(\Leftrightarrow\left[{}\begin{matrix}6x-3\pi=-2\pi+k24\pi.\\6x-3\pi=12\pi-2\pi+k24\pi.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k4\pi.\\x=\dfrac{13}{6}\pi+k4\pi.\end{matrix}\right.\)

Lời giải:

$\sin (\frac{x}{2}-\frac{\pi}{4})=\frac{-1}{2}=\sin (\frac{-\pi}{6})$

$\Rightarrow \frac{x}{2}-\frac{\pi}{4}=\frac{-\pi}{6}+2k\pi$ hoặc $\frac{x}{2}-\frac{\pi}{4}=\pi +\frac{\pi}{6}+2k\pi$ với $k$ nguyên 

$\Rightarrow x=\frac{\pi}{12}+4k\pi$ hoặc $x=\frac{17\pi}{6}+4k\pi$ với $k$ nguyên bất kỳ.

\(\Leftrightarrow\sin4x=-1\)

\(\Leftrightarrow4x=-\dfrac{\Pi}{2}+k2\Pi\)

hay \(x=-\dfrac{\Pi}{8}+\dfrac{k\Pi}{2}\)

a: \(\Leftrightarrow\sin3x=-\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=arcsin\left(-\dfrac{1}{3}\right)+k2\Pi\\3x=\Pi-arcsin\left(-\dfrac{1}{3}\right)+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(arcsin\left(-\dfrac{1}{3}\right)+k2\Pi\right)\\x=\dfrac{1}{3}\left(\Pi-arcsin\left(-\dfrac{1}{3}\right)+k2\Pi\right)\end{matrix}\right.\)

b: \(\Leftrightarrow-x+\dfrac{\Pi}{3}=k\Pi\)

hay \(x=-k\Pi+\dfrac{\Pi}{3}\)

\(\Leftrightarrow\sin3x=-\sin7x=\sin\left(-7x\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=-7x+k2\Pi\\3x=\Pi+7x+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{k\Pi}{5}\\x=-\dfrac{\Pi}{4}-\dfrac{k\Pi}{2}\end{matrix}\right.\)

\(\Leftrightarrow\sin3x=\sin x\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=x+k2\Pi\\3x=\Pi-x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\Pi\\x=\dfrac{\Pi}{4}+\dfrac{k\Pi}{2}\end{matrix}\right.\)

c: =>1/2*sin8x+1/2*sin2x=1/2*sin8x+1/2*sin4x

=>sin 4x=sin 2x

=>4x=2x+k2pi hoặc 4x=pi-2x+k2pi

=>x=kpi hoặc x=pi/6+kpi/3

e: =>(1+sin x)(1+cosx)=0

=>sin x=-1 hoặc cosx=-1

=>x=-pi/2+k2pi hoặc x=pi+k2pi

b.

\(\sqrt{3}cosx+sin2x=0\)

\(\Leftrightarrow cosx\left(\sqrt{3}+2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\\sqrt{3}+2sinx=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sinx=sin\left(\dfrac{-\pi}{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{3}+k2\pi\\x=\pi+\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}-x+2x=\dfrac{\Pi}{2}-\dfrac{\Pi}{4}+k2\Pi\\-x-2x=\dfrac{1}{2}\Pi+k2\Pi-\dfrac{\Pi}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{4}+k2\Pi\\x=-\dfrac{\Pi}{12}-\dfrac{k2\Pi}{3}\end{matrix}\right.\)