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\(1.sin3x+sin2x+sinx=cos2x+cosx+1\)
\(\Leftrightarrow2sin2x.cosx+sin2x=2cos^2x+cosx\)
\(\Leftrightarrow sin2x\left(2cosx+1\right)-cosx\left(2cosx+1\right)=0\\\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sin2x-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin2x=sin\left(\frac{\Pi}{2}-x\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\Pi}{3}+k2\Pi\\x=\frac{\Pi}{6}+m2\Pi orx=\frac{\Pi}{2}+k2\Pi\end{matrix}\right.\)
\(2.cos^2x+cos^23x=sin^22x\)
\(\Leftrightarrow2+cos2x+cos6x=1-cos4x\)
\(\Leftrightarrow1+cos2x+cos6x+cos4x=0\)
\(\Leftrightarrow2cos^2x+2cos5x.cosx=0\)
\(\Leftrightarrow2cosx\left(cosx+cos5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\Pi}{2}+k\Pi\\cos5x=cos\left(\Pi-x\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\Pi}{2}+k\Pi\\5x=\Pi-x+k2\Pi or5x=x-\Pi+k2\Pi\end{matrix}\right.\)
1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)
\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)
Do \(-1\le sin\left(2x+\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le1\)
b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)
\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)
\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)
c/ \(y=1-sin2x+2cos2x+\frac{3}{2}sin2x=\frac{1}{2}sin2x+2cos2x+1\)
\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)
\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)
Nhân 2 vế với \(sin4x\) sau đó tách:
\(\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=\frac{2sin2x.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}=\frac{4sinx.cosx.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}\)
Rồi rút gọn
đây là câu a
mk cảm thấy cứ hơi sai sai . bạn xem lại hộ mk nhé