1: \(\Leftrightarrow\sin^23x=\sin^2x\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin3x=\sin x\\\sin3x=\sin\left(-x\right)\end{matrix}\right.\)

TH1: sin 3x=sin x

\(\Leftrightarrow\left[{}\begin{matrix}3x=x+k2\Pi\\3x=\Pi-x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\Pi\\x=\dfrac{1}{4}\Pi+\dfrac{1}{2}k\Pi\end{matrix}\right.\)

5: \(\Leftrightarrow\Pi\cdot\sin\left(x-\dfrac{\Pi}{3}\right)=\dfrac{\Pi}{2}+k\Pi\)

\(\Leftrightarrow\sin\left(x-\dfrac{\Pi}{3}\right)=\dfrac{1}{2}+k\)

Để pt có nghiệm thì k+1/2>=-1 và k+1/2<=1

=>-3/2<=k<=1/2

Khi đó, pt sẽ có nghiệm là:

\(\left\{{}\begin{matrix}x-\dfrac{\Pi}{3}=arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\\x-\dfrac{\Pi}{3}=\Pi-arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi+\dfrac{\Pi}{3}\\x=\dfrac{4}{3}\Pi+arcsin\left(k+\dfrac{1}{2}\right)+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow1-2x=k2\Pi\)

hay \(x=\dfrac{1-k2\Pi}{2}\)

1 - 2x = k2II

hoặc x = \(\dfrac{1-k2II}{2}\)

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

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

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

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.

b: =>sin3x-1/2*sin2x+căn 3/2*cos2x=0

=>sin 3x=1/2*sin 2x-căn 3/2*cos2x=sin(2x-pi/3)

=>3x=2x-pi/3+k2pi hoặc 3x=4/3pi-2x+k2pi

=>x=-pi/3+k2pi hoặc x=4/15pi+k2pi/5

c: =>sin 2x-căn 3*cos2x=2cosx

=>sin(2x-pi/3)=cosx=sin(pi/2-x)

=>2x-pi/3=pi/2-x+k2pi hoặc 2x-pi/3=pi/2+x+k2pi

=>x=5/18pi+k2pi/3 hoặc x=5/6pi+k2pi

mn giúp em vs ạ

`sin x+4 cos x=2+2sin x`

`<=>sin x-4 cos x=-2`

`<=>\sqrt{17}(1/\sqrt{17} sin x-4/\sqrt{17} cos x)=-2`

`<=>1/\sqrt{17} sin x-4/\sqrt{17} cos x=-2/\sqrt{17}`

  Đặt `cos \alpha=1/\sqrt{17}` ; `sin \alpha= 4/\sqrt{17}`

  `=>cos \alpha sin x-sin \alpha cos x=-2/\sqrt{17}`

`<=>sin(x-\alpha)=-2/\sqrt{17}`

`<=>` $\left[\begin{matrix} x-\alpha=arc sin (\dfrac{-2}{\sqrt{17}})+k2\pi\\ x-\alpha=\pi-arc sin (\dfrac{-2}{\sqrt{17}})+k2\pi\end{matrix}\right.$ 

`<=>` $\left[\begin{matrix} x=\alpha+arc sin (\dfrac{-2}{\sqrt{17}})+k2\pi\\ x=\alpha+\pi-arc sin (\dfrac{-2}{\sqrt{17}})+k2\pi\end{matrix}\right.$   `(k in ZZ)`

Vậy `S={\alpha+arc sin ([-2]/\sqrt{17})+k2\pi,\alpha+\pi-arc sin ([-2]/\sqrt{17})+k2\pi|k in ZZ,cos \alpha=1/\sqrt{17}` ; `sin \alpha= 4/\sqrt{17}}`

ngưỡng mộ cj gkee:33

\(\dfrac{1}{sin2x}+\dfrac{1}{cos2x}=\dfrac{2}{sin4x}\)

\(ĐK:\left\{{}\begin{matrix}sin2x\ne0\\cos2x\ne0\end{matrix}\right.\)

         \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\) \(\left(k\in Z\right)\)

\(\Leftrightarrow\dfrac{sin2x+cos2x}{sin2x.cos2x}=\dfrac{2}{2\left(sin2x.cos2x\right)}\)

\(\Leftrightarrow\dfrac{sin2x+cos2x}{sin2x.cos2x}=\dfrac{1}{sin2x.cos2x}\)

\(\Leftrightarrow sin2x+cos2x=1\)

\(\Leftrightarrow\dfrac{1}{\sqrt{2}}sin2x+\dfrac{1}{\sqrt{2}}cos2x=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow sin2x.cos\dfrac{1}{4}\pi+cos2x.sin\dfrac{1}{4}\pi=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow sin\left(2x+\dfrac{1}{4}\pi\right)=sin\dfrac{1}{4}\pi\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{1}{4}\pi+k\pi\end{matrix}\right.\) `(tm)`