1) \(4cos^24x+2\left(\sqrt{3}+\sqrt{2}\right)cos4x+\sqrt{6}=0\)
2) \(cos4x+2+sin\left(2x+\frac{3\pi}{2}\right)=2cos^2x\)
3) \(sin\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin\left(\frac{\pi}{6}-x\right)=1\)
4) \(2cos\left(4x-\frac{\pi}{3}\right)+4cos2x=-1\)
5) \(cos^22x+cos^23x=sin^2x\)
6) \(sinx+\left(\sqrt{2}-1\right)cosx=1\)
7) \(cos2x-\left(\sqrt{3}+1\right)cosx+\frac{2+\sqrt{3}}{2}=0\)
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.
\(cos^22x+\frac{1}{2}+\frac{1}{2}cos6x=\frac{1}{2}-\frac{1}{2}cos2x\)
\(\Leftrightarrow cos^22x+\frac{1}{2}\left(cos6x+cos2x\right)=0\)
\(\Leftrightarrow cos^22x+cos4x.cos2x=0\)
\(\Leftrightarrow cos2x\left(cos2x+cos4x\right)=0\)
\(\Leftrightarrow cos2x\left(2cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
6.
\(\Leftrightarrow\frac{1}{\sqrt{4-2\sqrt{2}}}sinx+\frac{\sqrt{2}-1}{\sqrt{4-2\sqrt{2}}}cosx=\frac{1}{\sqrt{4-2\sqrt{2}}}\)
\(\Leftrightarrow sinx.cos\left(\frac{\pi}{8}\right)+cosx.sin\left(\frac{\pi}{8}\right)=cos\left(\frac{\pi}{8}\right)\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{8}\right)=sin\left(\frac{3\pi}{8}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{8}=\frac{3\pi}{8}+k2\pi\\x+\frac{\pi}{8}=\frac{5\pi}{8}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
7.
\(\Leftrightarrow2cos^2x-\left(\sqrt{3}+1\right)cosx+\frac{\sqrt{3}}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=-\frac{\sqrt{3}}{2}\end{matrix}\right.\) (bấm máy như bài 1)
\(\Leftrightarrow...\)
