1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=5x+k2\pi\\x-\dfrac{\pi}{3}=\pi-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

1.

ĐK: \(x\ne\dfrac{k\pi}{2}\)

\(cotx-tanx=sinx+cosx\)

\(\Leftrightarrow\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)

\(\Leftrightarrow\dfrac{cos^2x-sin^2x}{sinx.cosx}=sinx+cosx\)

\(\Leftrightarrow\left(\dfrac{cosx-sinx}{sinx.cosx}-1\right)\left(sinx+cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx=sinx.cosx\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)

\(\left(2\right)\Leftrightarrow t=\dfrac{1-t^2}{2}\left(t=cosx-sinx,\left|t\right|\le2\right)\)

\(\Leftrightarrow t^2+2t-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow cosx-sinx=-1+\sqrt{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)

1.

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(\left(m+1\right)^2+\left(-3\right)^2\ge m^2\)

\(\Leftrightarrow...\)

2.

\(\Leftrightarrow3\left(\frac{1}{2}-\frac{1}{2}cos2x\right)+4m.sin2x-4=0\)

\(\Leftrightarrow8m.sin2x-3cos2x=5\)

Pt vô nghiệm khi: \(\left(8m\right)^2+\left(-3\right)^2< 5^2\)

\(\Leftrightarrow...\)

Điều kiện xác định:

\(sinx+\sqrt{3}cosx\ge0\Leftrightarrow tanx\ge-\sqrt{3}\Leftrightarrow x\ge\dfrac{2\pi}{3}+k\pi\)

Đặt \(t=\sqrt{sinx+\sqrt{3}cosx},t\ge0\)

Phương trình đã cho trở thành:

\(t^2+t-2=0\Leftrightarrow\left(t-1\right)\left(t+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(\text{nhận}\right)\\t=-2\left(\text{loại}\right)\end{matrix}\right.\)

Với t = 1, ta có

\(sinx+\sqrt{3}cosx=1\Leftrightarrow2.\left(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right)=1\)

\(\Leftrightarrow2.cos\left(x-\dfrac{\pi}{6}\right)=1\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

Đối chiếu với điều kiện xác định, ta phải có

\(\left\{{}\begin{matrix}\dfrac{\pi}{2}+k2\pi\ge\dfrac{2\pi}{3}+k\pi\\-\dfrac{\pi}{6}+k2\pi\ge\dfrac{2\pi}{3}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}k\ge\dfrac{1}{6}\\k\ge\dfrac{5}{6}\end{matrix}\right.\) \(\Rightarrow k\ge1\)

Vậy phương trình có hai họ nghiệm là \(x=\dfrac{\pi}{2}+k2\pi\) và \(x=-\dfrac{\pi}{6}+k2\pi\) với \(k\in Z,k\ge1\)

Lời giải:

PT $\Leftrightarrow 2\sin 2x\cos 2x+2\cos 2x+4(\sin x+\cos x)=1+\cos ^22x-\sin ^22x=2\cos ^22x$

$\Leftrightarrow \sin 2x\cos 2x+\cos 2x+2(\sin x+\cos x)=\cos ^22x$

$\Leftrightarrow \cos 2x(\sin 2x+1-\cos 2x)+2(\sin x+\cos x)=0$

$\Leftrightarrow \cos 2x(2\sin x\cos x+2\sin ^2x)+2(\sin x+\cos x)=0$

$\Leftrightarrow \cos 2x\sin x(\cos x+\sin x)+(\sin x+\cos x)=0$

$\Leftrightarrow (\sin x+\cos x)(\cos 2x\sin x+1)=0$

Nếu $\sin x+\cos x=0$. Kết hợp $\sin ^2x+\cos ^2x=1$ suy ra $(\sin x, \cos x)=(\frac{1}{\sqrt{2}}; \frac{-1}{\sqrt{2}})$ và hoán vị

$\Rightarrow x=k\pi -\frac{\pi}{4}$ với $k$ nguyên.

Nếu $\cos 2x\sin x+1=0$

$\Leftrightarrow (1-2\sin ^2x)\sin x+1=0$

$\Leftrightarrow (1-\sin x)(2\sin ^2x+2\sin x+1)=0$

$\Rightarrow \sin x=1$

$\Rightarrow x=2k\pi +\frac{\pi}{2}$ với $k$ nguyên.