Bài 2: Phương trình lượng giác cơ bản

\(y=2cos^2x-\sqrt{3}sin2x+cos^2x+sin^2x=cos2x-\sqrt{3}sin2x+2\)

\(=2\left(\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x\right)+2=2cos\left(2x+\frac{\pi}{3}\right)+2\)

\(\Rightarrow0\le y\le4\)

\(cosx=sin\frac{\pi}{3}\)

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

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

\(\Leftrightarrow cosx=cos\left(\frac{\pi}{6}\right)\)

\(\Rightarrow x=\pm\frac{\pi}{6}+k2\pi\)

\(\left(m+1\right)\left(\frac{1-cos2x}{2}\right)-sin2x+\frac{1+cos2x}{2}=0\)

\(\Leftrightarrow m+2-mcos2x-2sin2x=0\)

\(\Leftrightarrow2sin2x+m.cos2x=m+2\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

\(2^2+m^2\ge\left(m+2\right)^2\)

\(\Leftrightarrow m\le0\)

Đặt \(tan\frac{x}{2}=t\Rightarrow t\ne0\)

\(\Rightarrow\left(2m-1\right)t+m=0\) có nghiệm \(t\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\2m-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\m\ne\frac{1}{2}\end{matrix}\right.\)

Lời giải:

$\sin (-2x+3)=\frac{-\sqrt{3}}{2}$

\(\Rightarrow \left[\begin{matrix} -2x+3=\frac{-\pi}{3}+2k\pi \\ -2x+3=\frac{4}{3}\pi +2k\pi \end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{\pi}{6}+\frac{3}{2}-k\pi\\ x=\frac{-2}{3}\pi +\frac{3}{2}-k\pi \end{matrix}\right.\) với $k$ nguyên.

Phương trình lượng giác bậc nhất cơ bản mà :(

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{12}+k2\pi\\x=\frac{13}{12}\pi+k2\pi\end{matrix}\right.\)

\(th1:0\le\frac{5\pi}{12}+k2\pi\le2\pi\)

\(th2:0\le\frac{13}{12}\pi+k2\pi\le2\pi\)

Chặn k là okie :)

\(\Leftrightarrow x+10^0=-90^0+k360^0\)

\(\Leftrightarrow x=-100^0+k360^0\)

ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\)

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

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

\(\Leftrightarrow\frac{sin2x.cosx}{cosx}=\sqrt{2}sinx\)

\(\Leftrightarrow2sinx.cosx=\sqrt{2}sinx\)

\(\Leftrightarrow cosx=\frac{\sqrt{2}}{2}\)

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

Vậy \(x=-\frac{\pi}{4}+k2\pi\)