1.

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-\frac{\sqrt{3}}{2}\\cos4x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow x=...\)

(Cứ bấm máy giải pt bậc 2 như bt, nó cho 2 nghiệm rất xấu, bạn lưu 2 nghiệm vào 2 biến A; B rồi thoát ra ngoài MODE-1, tính \(\sqrt{A^2}\) và \(\sqrt{B^2}\) sẽ ra dạng căn đẹp của 2 nghiệm, lưu ý dấu so với nghiệm ban đầu)

2.

\(\Leftrightarrow cos4x+1+sin\left(2x-\frac{\pi}{2}\right)=cos2x\)

\(\Leftrightarrow2cos^22x-cos2x=cos2x\)

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

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

3.

\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]=\frac{1}{2}\)

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

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x+\frac{2\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow...\)

4.

\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)

\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)

\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)

\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)

\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)

\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)

5.

\(cos^22x+\frac{1}{2}+\frac{1}{2}cos6x=\frac{1}{2}-\frac{1}{2}cos2x\)

\(\Leftrightarrow cos^22x+\frac{1}{2}\left(cos6x+cos2x\right)=0\)

\(\Leftrightarrow cos^22x+cos4x.cos2x=0\)

\(\Leftrightarrow cos2x\left(cos2x+cos4x\right)=0\)

\(\Leftrightarrow cos2x\left(2cos^22x+cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

2a.

\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)

\(\Leftrightarrow cos^2x+2cosx+1=0\)

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

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

a/

\(\Leftrightarrow4cos^2\left(6x-2\right)+8\left(1+cos\left(6x-2\right)\right)-13=0\)

Đặt \(cos\left(6x-2\right)=a\Rightarrow\left|a\right|\le1\)

Pt trở thành:

\(4a^2+8\left(1+a\right)-13=0\)

\(\Leftrightarrow4a^2+8a-5=0\Rightarrow\left[{}\begin{matrix}a=\frac{1}{2}\\a=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cos\left(6x-2\right)=\frac{1}{2}\)

\(\Rightarrow6x-2=\pm\frac{\pi}{3}+k2\pi\)

\(\Rightarrow x=\frac{1}{3}\pm\frac{\pi}{18}+\frac{k\pi}{3}\)

b/

\(\Leftrightarrow2cos^2\left(x+75^0\right)-1+3sin\left(15^0-x\right)-1=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(90^0-15^0+x\right)-2=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(x+75^0\right)-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cos\left(x+75^0\right)=\frac{1}{2}\\cos\left(x+75^0\right)=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+75^0=60^0+k360^0\\x+75^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+k360^0\\x=-135^0+k360^0\end{matrix}\right.\)

c/

\(\Leftrightarrow\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x\right)+\left(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx\right)=1\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow cos2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow1-2sin^2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)\left(1-2sin\left(x-\frac{\pi}{6}\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{6}\right)=0\\sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)

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

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

a/

\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)

\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)

\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)

\(\Leftrightarrow-sin4x=sin^24x\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)

\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)

\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

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

1.

\(\Leftrightarrow3sin^22x+1-sin^22x=3sin2x\)

\(\Leftrightarrow2sin^22x-3sin2x+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=\frac{1}{2}\end{matrix}\right.\)

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

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

b/

\(\Leftrightarrow3cos^2x+4sin\left(2\pi-\frac{\pi}{2}-x\right)+1=0\)

\(\Leftrightarrow3cos^2x-4sin\left(x+\frac{\pi}{2}\right)+1=0\)

\(\Leftrightarrow3cos^2x-4cosx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm arcos\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow1-sin^22x+\sqrt{3}sin2x+sin2x=1+\sqrt{3}\)

\(\Leftrightarrow-sin^22x+\left(\sqrt{3}+1\right)sin2x-\sqrt{3}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=\sqrt{3}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)

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

d/

\(\Leftrightarrow4\left(1-2sin^2x\right)+5sinx=4\left(3sinx-4sin^3x\right)+5\)

\(\Leftrightarrow16sin^3x-8sin^2x-7sinx-1=0\)

\(\Leftrightarrow\left(sinx-1\right)\left(4sinx+1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\frac{1}{4}\end{matrix}\right.\)

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

a

\(\Leftrightarrow\left(3sinx-sin3x\right)cos3x+\left(3cosx+cos3x\right)sin3x+3\sqrt{3}cos4x=3\)

\(\Leftrightarrow\left(sinx.cos3x+sin3x.cosx\right)+\sqrt{3}cos4x=1\)

\(\Leftrightarrow sin4x+\sqrt{3}cos4x=1\)

Tới đây thôi, mình lười ghi rồi =))

b

\(\Leftrightarrow\left(1-cos2x\right)\left(2sin^2x-1\right)\left(2sin^2+1\right)=cos2x\left(7cos^22x+3cos2x-4\right)\)

\(\Leftrightarrow\left(1-cos2x\right)\left(-cos2x\right)\left(2-cos2x\right)=cos2x\left(7cos^22x+3cos2x+4\right)\)

\(\Leftrightarrow-cos^22x+3cos2x-2=7cos^22x+3cos2x+4\)

\(\Leftrightarrow4cos^22x+3=0\)

=> pt vô nghiệm