a.
Ta có: \(\left\{{}\begin{matrix}sin^4x\le sin^2x\\cos^5x\le cos^2x\end{matrix}\right.\)
\(\Rightarrow sin^4x+cos^5x\le sin^2x+cos^2x=1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sin^4x=sin^2x\\cos^5x=cos^2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}cosx=1\\sinx=0\end{matrix}\right.\\\left\{{}\begin{matrix}sin^2x=1\\cosx=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
b.
\(sinx.sin\left(2x+\frac{\pi}{6}\right)=cos\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{6}\right)-cos\left(3x+\frac{\pi}{6}\right)=2cos\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=-cos\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=cos\left(\frac{5\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{6}=\frac{5\pi}{6}-x+k2\pi\\3x+\frac{\pi}{6}=x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
c.
Nhận thấy \(sinx=0\) ko phải nghiệm
\(\Leftrightarrow8sinx.cosx.cos2x.cos4x=sinx\)
\(\Leftrightarrow4sin2x.cos2x.cos4x=sinx\)
\(\Leftrightarrow2sin4x.cos4x=sinx\)
\(\Leftrightarrow sin8x=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=x+k2\pi\\8x=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{7}\\x=\frac{\pi}{9}+\frac{k2\pi}{9}\end{matrix}\right.\)
d.
\(\Leftrightarrow2sinx.cosx-2cosx=\sqrt{3}cos2x\)
\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)
\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=cosx\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
e.
\(\Leftrightarrow sin^2x+2sinx.cosx+3cos^2x=3\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(tan^2x+2tanx+3=3\left(1+tan^2x\right)\)
\(\Leftrightarrow2tan^2x-2tanx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=0\\tanx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
f.
\(\Leftrightarrow cos2x-cos4x+2cos^22x=sin2x\)
\(\Leftrightarrow cos2x-cos4x+cos4x+1=sin2x\)
\(\Leftrightarrow sin2x-cos2x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
g.
\(\Leftrightarrow2cos4x.cos2x-2cos4x.sin2x=2cos4x\)
\(\Leftrightarrow cos4x\left(cos2x-sin2x-1\right)=0\)
\(\Leftrightarrow cos4x\left[\sqrt{2}cos\left(2x+\frac{\pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
