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

\(\Leftrightarrow\left[{}\begin{matrix}1=-1+k2\pi\left(vn\right)\\x=\frac{k\pi}{2}\end{matrix}\right.\)

Vậy nghiệm của pt là \(x=\frac{k\pi}{2}\)

\(\sqrt{3}cosx+2sin^2\left(\dfrac{x}{2}-\pi\right)=1\) 

\(\Leftrightarrow\sqrt{3}cosx+2sin^2\dfrac{x}{2}=1\)

\(\Leftrightarrow\sqrt{3}cosx-cosx=0\Leftrightarrow cosx=0\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) ( k thuộc Z )

Vậy ... 

22.

Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)

\(3tan^2x+2tanx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\dfrac{1}{3}\end{matrix}\right.\)

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

Nghiệm dương nhỏ nhất của pt là: \(x=arctan\left(\dfrac{1}{3}\right)\)

22. PT đã cho tương đương

3 - 4cos²x + 2 sinxcosx = 0

⇔ 3 - 2 - 2cos2x + sin2x = 0

⇔ 1 - 2cos2x + sin2x = 0

⇔ 1 + sin2x = 2cos2x

⇔ sin\(\dfrac{\pi}{2}\) + sin2x = 2cos2x

⇔ \(2sin\left(\dfrac{\pi}{4}+x\right).cos\left(\dfrac{\pi}{4}-x\right)\) = 2cos2x

Do \(\left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}+x\right)=\dfrac{\pi}{2}\) 

⇒ \(sin\left(\dfrac{\pi}{4}+x\right)=cos\left(\dfrac{\pi}{4}-x\right)\)

Vậy sin²\(\left(x+\dfrac{\pi}{4}\right)\) = cos2x

Cái này là hiển nhiên ????

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{2}+k2\pi\\x\ne\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\\end{matrix}\right.\)

\(\dfrac{cosx-2sinx.cosx}{2cos^2x-1-sinx}=\sqrt{3}\)

\(\Leftrightarrow\dfrac{cosx-sin2x}{cos2x-sinx}=\sqrt{3}\)

\(\Rightarrow cosx-sin2x=\sqrt{3}cos2x-\sqrt{3}sinx\)

\(\Leftrightarrow cosx+\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

\(\Leftrightarrow\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)

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

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

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

Vậy \(x=-\dfrac{\pi}{6}+k2\pi\)

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

\(\Leftrightarrow sin^2x.\dfrac{1}{2}+2.sinx.cosx-\dfrac{5}{2}cos^2x=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+5cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\sinx=-5cosx\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-5\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arc.tan\left(-5\right)+k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

ĐKXĐ: \(x\ne-\frac{\pi}{4}+k\pi\)

\(\Leftrightarrow\frac{\left(1-sin^2x\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1+sinx\right)\)

\(\Leftrightarrow\frac{\left(1+sinx\right)\left(1-sinx\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1+sinx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\\\frac{\left(1-sinx\right)\left(cosx-1\right)}{sinx+cosx}=2\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow cosx-1-sinx.cosx+sinx=2sinx+2cosx\)

\(\Leftrightarrow sinx+cosx+sinx.cosx+1=0\)

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

\(\Leftrightarrow...\)

1.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos4x=\dfrac{1}{2}+\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow-cos4x=cos\left(2x-\dfrac{\pi}{2}\right)\)

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

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

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

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{3}\)

2.

\(\Leftrightarrow1-cos^2x+1-sin^24x=2\)

\(\Leftrightarrow cos^2x+sin^24x=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}cosx=0\\sin4x=0\end{matrix}\right.\)

\(\Leftrightarrow cosx=0\)

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

1, \(sin\left(x+\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{6}}{2}\)

⇔  \(\dfrac{\sqrt{2}}{2}sin\left(x+\dfrac{\pi}{6}\right)+\dfrac{\sqrt{2}}{2}cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

⇔ \(sin\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)

2, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx=1-\sqrt{3}\)

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

⇔ sinx . si

cái này trên OLM mà

cái này chắc cũng lớp 10 chứ ko thoát đâu

nhưng nếu chỗ =căn 3+1 bỏ căn 3 thì tui ms làm đc

a/

\(\Leftrightarrow cos\frac{4x}{3}=\frac{cos2x+1}{2}\)

Đặt \(\frac{2x}{3}=a\Rightarrow2x=3a\)

Pt trở thành:

\(cos2a=\frac{cos3a+1}{2}\)

\(\Leftrightarrow2\left(2cos^2a-1\right)=4cos^3a-3cosa+1\)

\(\Leftrightarrow4cos^3a-4cos^2a-3cosa+3=0\)

\(\Leftrightarrow\left(cosa-1\right)\left(4cos^2a-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosa=1\\cosa=\frac{\sqrt{3}}{2}\\cosa=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\frac{2x}{3}\right)=1\\cos\left(\frac{2x}{3}\right)=\frac{\sqrt{3}}{2}\\cos\left(\frac{2x}{3}\right)=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2x}{3}=k2\pi\\\frac{2x}{3}=\pm\frac{\pi}{6}+k2\pi\\\frac{2x}{3}=\pm\frac{7\pi}{6}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\)

b/

Đặt \(\frac{2x}{3}=a\)

\(\Rightarrow cos4a=cos^2a\)

\(\Leftrightarrow2cos^22a-1=\frac{1+cos2a}{2}\)

\(\Leftrightarrow4cos^22a-cos2a-3=0\)

\(\Rightarrow\left[{}\begin{matrix}cos2a=1\\cos2a=-\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cos\left(\frac{4x}{3}\right)=1\\cos\left(\frac{4x}{3}\right)=-\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{4x}{3}=k2\pi\\\frac{4x}{3}=\pm arccos\left(-\frac{3}{4}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k3\pi}{2}\\x=\pm\frac{3}{4}arccos\left(-\frac{3}{4}\right)+\frac{k3\pi}{2}\end{matrix}\right.\)

c/

\(\Leftrightarrow cos\frac{6x}{5}+2=3cos\frac{4x}{5}\)

Đặt \(\frac{2x}{5}=a\)

\(\Rightarrow cos3a+2=3cos2a\)

\(\Leftrightarrow4cos^3a-3cosa+2=6cos^2a-3\)

\(\Leftrightarrow4cos^3a-6cos^2a-3cosa+5=0\)

\(\Leftrightarrow\left(cosa-1\right)\left(4cos^2a-2cosa-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosa=1\\cosa=\frac{1+\sqrt{21}}{4}>1\left(l\right)\\cosa=\frac{1-\sqrt{21}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cos\left(\frac{2x}{5}\right)=1\\cos\left(\frac{2x}{5}\right)=\frac{1-\sqrt{21}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2x}{5}=k2\pi\\\frac{2x}{5}=\pm arccos\left(\frac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k5\pi\\x=\pm\frac{5}{2}arccos\left(\frac{1-\sqrt{21}}{4}\right)+k5\pi\end{matrix}\right.\)

a) \(cos\left(4x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\Rightarrow cos\left(4x+\dfrac{\pi}{3}\right)=cos\dfrac{\pi}{6}\)

                                      \(\Rightarrow\left[{}\begin{matrix}4x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\4x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

                                            ..... bạn tự tìm x nhé!

b)\(sin^2x-3sin3x+2=0\)\(\Rightarrow sin^2x-3\left(3sinx-4sin^3x\right)+2=0\)

\(\Rightarrow12sin^3x+sin^2x-9sinx+2=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\dfrac{2}{3}\\sinx=\dfrac{1}{4}\end{matrix}\right.\)    \(\Rightarrow\).... bạn tự tìm x nhé!

c)\(tan\left(2x+10^o\right)=\sqrt{3}\Rightarrow tan\left(2x+10^o\right)=tan60^o\)

                                     \(\Rightarrow2x+10^o=60^o+k180^o\)

                                     \(\Rightarrow x=25^o+k90^o\left(k\in Z\right)\)

d) \(tanx\cdot cot2x=1\)

Đk: \(\left\{{}\begin{matrix}cosx\ne0\\sin2x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m\pi\\x\ne m\dfrac{\pi}{2}\end{matrix}\right.\)

Pt: \(\Rightarrow tanx=tan2x\Rightarrow x=2x+k\pi\)

                                 \(\Rightarrow x=k\pi\)

  Đối chiếu với đk trên thỏa mãn đk\(\Rightarrow x=k\pi\)