1. \(\sin^2x+\sin2x=3\cos^2x\Leftrightarrow\sin^2x+2\sin x\cos x-3\cos^2x=0\Leftrightarrow4\sin^2x+2\sin x\cos x-3=0\)

Vì \(\cos x=0\) không phải là nghiệm của phương trình, nên chia 2 vế pt cho \(\cos x\), ta đc:

\(4\tan^2x+2\tan x-\frac{3}{\cos^2x}=0\Leftrightarrow4\tan^2x+2\tan x-3\left(1+\tan^2x\right)=0\Leftrightarrow\tan^2x+2\tan x-3=0\)

Suy ra: \(\begin{matrix}\tan x=1\\\tan x=-3\end{matrix}\) suy ra x.

b) \(\Leftrightarrow\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\sin2x\Leftrightarrow\sin\left(x+\frac{\pi}{4}\right)=\sin2x\Leftrightarrow\begin{cases}x+\frac{\pi}{4}=2x+k2\pi\\x+\frac{\pi}{4}=\pi-2x+k2\pi\end{cases}\)

\(\Leftrightarrow\begin{cases}x=\frac{\pi}{4}-k2\pi\\x=\frac{\pi}{4}+\frac{k2\pi}{3}\end{cases}\)

Vậy ....

Chỗ Viết các nghiệm: Sửa lại : dùng dấu  ngoặc vuông thay cho ngoặc nhọn

3.

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

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)

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

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

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(sin\left(sinx\right)=0\)

\(\Leftrightarrow sinx=k\pi\) (1)

Do \(-1\le sinx\le1\Rightarrow-1\le k\pi\le1\)

\(\Rightarrow-\dfrac{1}{\pi}\le k\le\dfrac{1}{\pi}\Rightarrow k=0\) do \(k\in Z\)

Thế vào (1)

\(\Rightarrow sinx=0\Rightarrow x=n\pi\)

2.

\(sin\left(cosx\right)=1\Leftrightarrow cosx=\dfrac{\pi}{2}+k2\pi\)

Do \(-1\le cosx\le1\Rightarrow-1\le\dfrac{\pi}{2}+k2\pi\le1\)

\(\Rightarrow-\dfrac{1}{2\pi}-\dfrac{1}{4}\le k\le\dfrac{1}{2\pi}-\dfrac{1}{4}\) 

\(\Rightarrow\) Không tồn tại k thỏa mãn

Pt vô nghiệm

ĐK: \(x\ne\dfrac{\pi}{6}+k2\pi;x\ne\dfrac{5\pi}{6}+k2\pi\)

\(\dfrac{cosx-\sqrt{3}sinx}{sinx-\dfrac{1}{2}}=0\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=0\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=0\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

Đối chiếu điều kiện ta được \(x=-\dfrac{5\pi}{6}+k2\pi\).

1.

\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)

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

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

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

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

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

2.

\(\left|cosx-sinx\right|+2sin2x=1\)

\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)

\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)

\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)

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

3.

\(2sin2x-3\sqrt{6}\left|sinx+cosx\right|+8=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^2-3\sqrt{6}\left|sinx+cosx\right|+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|sinx+cosx\right|=\sqrt{6}\left(vn\right)\\\left|sinx+cosx\right|=\dfrac{\sqrt{6}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left|sin\left(x+\dfrac{\pi}{4}\right)\right|=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\pm\dfrac{\sqrt{3}}{2}\)

...

2.1

a.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

b.

\(cosx-\sqrt{3}sinx=1\)

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

c.

\(\sqrt{3}sin\dfrac{x}{3}+cos\dfrac{x}{2}=\sqrt{2}\)

Câu này đề đúng không nhỉ? Nhìn thấy có vẻ không đúng lắm

d.

\(cosx-sinx=1\)

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

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

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

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

\(sin^3x+cos^3x-sinx-cosx=cos2x\)

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

​​\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)-\left(sinx+cosx\right)-\left(cosx+sinx\right)\left(cosx-sinx\right)=0\)​​

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

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

TH1: (1)\(\Leftrightarrow\sqrt{2}.sin\left(x+\dfrac{\pi}{4}\right)=0\)\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

TH2: Đặt \(t=sinx-cosx\) ;\(t\in\left(-2;2\right)\)

\(\Rightarrow\dfrac{t^2-1}{2}=-sinx.cosx\)

Pt (2)\(\Rightarrow t+\dfrac{t^2-1}{2}=0\)\(\Leftrightarrow t^2+2t-1=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\left(tm\right)\\t=-1-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=-1+\sqrt{2}\)\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}+1\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arc.cos\dfrac{1-\sqrt{2}}{2}+k2\pi\\x=\dfrac{-\pi}{4}-arc.cos\dfrac{1-\sqrt{2}}{2}+k2\pi\end{matrix}\right.\)(\(k\in\)\(Z\))

Vậy...

\(sin^23x.cos2x+sin^2x=0\)

\(\left(3sinx-4sin^3x\right)^2.cos2x+sin^2x=0\)

\(sin^2x\left[\left(3-4sin^2x\right)^2.cos2x+1\right]=0\)

\(sin^2x\left[\left(1+2cos2x\right)^2.cos2x+1\right]=0\)

\(sin^2x\left(4cos^22x+1\right)\left(cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\text{π}\\2x=k2\text{π}\end{matrix}\right.\)\(\Leftrightarrow x=k\text{π}\)

\(sin^23xcos2x+sin^2x=0\rightarrow\dfrac{1-cos6x}{2}.cos2x+\dfrac{1-cos2x}{2}=0\\ \rightarrow cos6xcos2x=1\rightarrow cos8x+cos4x=2\\ \rightarrow cos8x=cos4x=1\rightarrow x=\dfrac{k\pi}{2}\left(k\in Z\right)\)