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1.
\(\Leftrightarrow2cos2x+sinx-sin3x=0\)
\(\Leftrightarrow2cos2x-2cos2x.sinx=0\)
\(\Leftrightarrow2cos2x\left(1-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
2.
\(cos^2x+\left(sin3x-1\right)\left(1-cos\left(\dfrac{\pi}{2}-x\right)\right)=0\)
\(\Leftrightarrow1-sin^2x+\left(sin3x-1\right)\left(1-sinx\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx\right)+\left(sin3x-1\right)\left(1-sinx\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx+sin3x-1\right)=0\)
\(\Leftrightarrow2\left(1-sinx\right)sin2x.cosx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sin2x=0\\cosx=0\end{matrix}\right.\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow x=\dfrac{k\pi}{2}\)
1.
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)
\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)
Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)
\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)
2.
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)
\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
1.
\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\)
\(\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{5}}{3}\)
\(tanx=\dfrac{sinx}{cosx}=\dfrac{2}{\sqrt{5}}\)
\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sinx+cosx\right)=\dfrac{\sqrt{10}+2\sqrt{2}}{6}\)
2.
Đề bài thiếu, cos?x
Và x thuộc khoảng nào?
3.
\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow sinx;cosx>0\)
\(\dfrac{1}{cos^2x}=1+tan^2x=5\Rightarrow cos^2x=\dfrac{1}{5}\Rightarrow cosx=\dfrac{\sqrt{5}}{5}\)
\(sinx=cosx.tanx=\dfrac{2\sqrt{5}}{5}\)
4.
\(A=\left(2cos^2x-1\right)-2cos^2x+sinx+1=sinx\)
\(B=\dfrac{cos3x+cosx+cos2x}{cos2x}=\dfrac{2cos2x.cosx+cos2x}{cos2x}=\dfrac{cos2x\left(2cosx+1\right)}{cos2x}=2cosx+1\)
=>2cos2x=pi(loại) hoặc sin x-cosx=0
=>sin x-cosx=0
=>sin(x-pi/4)=0
=>x-pi/4=kpi
=>x=kpi+pi/4
mà x\(\in\left[-pi;pi\right]\)
nên \(x\in\left\{\dfrac{pi}{4};-\dfrac{3}{4}pi\right\}\)
=> D
Bạn sai ở chỗ này:
\(2cos2x=2cos2x.sinx\)
\(\Leftrightarrow sinx=\frac{2cos2x}{2cos2x}\)
Đúng ra phải là: \(\Leftrightarrow2cos2x.sinx-2cos2x=0\)
\(\Leftrightarrow2cos2x\left(sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=1\end{matrix}\right.\)