\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=2cos2x\)

\(\Leftrightarrow1-\frac{1}{2}sin^22x=2cos2x\)

\(\Leftrightarrow2-\left(1-cos^22x\right)=4cos2x\)

\(\Leftrightarrow cos^22x-4cos2x+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=2+\sqrt{3}>1\left(l\right)\\cos2x=2-\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\frac{1}{2}arccos\left(2-\sqrt{3}\right)+k\pi\)

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.

P/t \(\Leftrightarrow2cos2x.sin2x-sin2x+2cos^22x-cos2x-1=0\)

\(\Leftrightarrow sin4x-sin2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sinx.cos3x-2sin3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos3x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(1\right)\\cos3x=sin3x\left(2\right)\end{matrix}\right.\) 

(1) \(\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) \(\Leftrightarrow sin3x-cos3x=0\)  \(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\left(k\in Z\right)\)

Vậy ... 

ta có : \(VT=\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}=\dfrac{1-sin2x}{1+sin2x}=\dfrac{sin^2x-2sinx.cosx+cos^2x}{sin^2x+2sinx.cosx+cos^2x}\)

\(=\left(\dfrac{sinx-cosx}{sinx+cosx}\right)^2=\left(\dfrac{\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)}\right)=tan^2\left(x-\dfrac{\pi}{4}\right)\)

\(=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\left(đpcm\right)\)

ta có : \(VT=\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}=\dfrac{1-sin2x}{1+sin2x}=\dfrac{sin^2x-2sinx.cosx+cos^2x}{sin^2x+2sinx.cosx+cos^2x}\)

\(=\left(\dfrac{sinx-cosx}{sinx+cosx}\right)^2=\left(\dfrac{\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)}\right)=tan^2\left(x-\dfrac{\pi}{4}\right)\)

\(=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\left(đpcm\right)\)

ta có : \(VT=\dfrac{2cos2x-sin4x}{2cos2x+sin4x}=\dfrac{2cos2x-2sin2x.cos2x}{2cos2x+2sin2x.cos2x}\)

\(=\dfrac{2cos2x\left(1-sin2x\right)}{2cos2x\left(1+sin2x\right)}=\dfrac{1-sin2x}{1+sin2x}=\dfrac{sin^2x-2sinx.cosx+cos^2x}{sin^2x+2sinx.cosx+cos^2x}\)

\(=\left(\dfrac{sinx-cosx}{sinx+cosx}\right)^2=\left(\dfrac{\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)}\right)=tan^2\left(x-\dfrac{\pi}{4}\right)\)

\(=tan^2\left(\dfrac{\pi}{4}-x\right)=VP\left(đpcm\right)\)

\(pt\Leftrightarrow sin2x=-2cosx\\ \text{Mà }sin^22x+cos^22x=1\\ \Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{1}{5}\\cos2x=-\frac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{arccos\left(\frac{1}{5}\right)}{2}+m\pi\\x=\pm\frac{arccos\left(-\frac{1}{5}\right)}{2}+n\pi\end{matrix}\right.\)

Có 2 cách giải bài này:

Cách 1.

Nhận thấy \(cos2x=0\) không phải nghiệm, chia 2 vế cho \(cos2x\) ta được:

\(2+\frac{sin2x}{cos2x}=0\Leftrightarrow2+tan2x=0\Rightarrow tan2x=-2\)

Đặt \(tana=-2\Rightarrow tan2x=tana\)

\(\Rightarrow2x=a+k\pi\Rightarrow x=\frac{a}{2}+\frac{k\pi}{2}\)

(Hoặc sử dụng trực tiếp \(2x=arctan\left(-2\right)+k\pi\Rightarrow x=\frac{arctan\left(-2\right)}{2}+\frac{k\pi}{2}\))

Cách 2:

Với dạng \(a.sint+b.cost=c\) thì cách giải chung là chia 2 vế cho \(\sqrt{a^2+b^2}\) , khi đó 2 hệ số \(\frac{a}{\sqrt{a^2+b^2}}\) và \(\frac{b}{\sqrt{a^2+b^2}}\) có tổng bình phương bằng 1 nên có thể đặt thành sin, cos và sử dụng công thức lượng giác

Chia 2 vế cho \(\sqrt{5}\) ta được:

\(\frac{1}{\sqrt{5}}sin2x+\frac{2}{\sqrt{5}}cos2x=0\) (để ý rằng \(\left(\frac{1}{\sqrt{5}}\right)^2+\left(\frac{2}{\sqrt{5}}\right)^2=1\) là 1 tính chất cơ bản của sin, cos)

Đặt \(\left\{{}\begin{matrix}\frac{1}{\sqrt{5}}=cosa\\\frac{2}{\sqrt{5}}=sina\end{matrix}\right.\) ta được

\(sin2x.sina+cos2x.cosa=0\)

\(\Leftrightarrow sin\left(2x+a\right)=0\)

\(\Rightarrow2x+a=k\pi\Rightarrow x=-\frac{a}{2}+\frac{k\pi}{2}\)

1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

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

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)