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điều kiện : cosx\(\ne\)\(\frac{1}{\sqrt{2}}\)=> x\(\ne\)\(\pm\)\(\frac{\pi}{4}\)+2k\(\pi\), k\(\in\)Z
pt<=> tử số =0
<=>cos2x-sin(3x-\(\frac{\pi}{4}\)+x+\(\frac{3\pi}{4}\))-sin(3x-\(\frac{\pi}{4}\)-x-\(\frac{3\pi}{4}\))-2=0
<=> cos2x-sin(x+\(\frac{\pi}{2}\))-sin(2x-\(\pi\))-2=0
<=> cos2x-cosx+sin2x-2sin2x-2cos2x=0
<=>-cos2x-coxs+2sinx.cosx-2sin2x=0
đến đây bạn nhóm lại ra nghiệm rồi kiểm tra đk là xong
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.
\(cos^2x+\left(sin3x-1\right)\left(1-cos\left(\dfrac{\pi}{2}-x\right)\right)=0\)
\(\Leftrightarrow1-sin^2x+\left(sin3x-1\right)\left(1-sinx\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx\right)+\left(sin3x-1\right)\left(1-sinx\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx+sin3x-1\right)=0\)
\(\Leftrightarrow2\left(1-sinx\right)sin2x.cosx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sin2x=0\\cosx=0\end{matrix}\right.\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow x=\dfrac{k\pi}{2}\)
D=sin(pi+x)+sinx+cot(pi-x)+tan(pi/2-x)
=-sinx+sinx-cotx+cotx=0
phương trình tương đương:
sin2x+cos2x+\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))+2sinx.cosx+cos2x-sin2x=0
<=> 2cos2x+2sinx.cosx+\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))=0
<=> 2cosx(sinx+cosx)+\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))=0
<=>(2cosx+1).\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))=0
<=>\(\left[\begin{array}{nghiempt}sin\left(x+\frac{\pi}{4}\right)=0\\2cosx+1=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=k\pi-\frac{\pi}{4}\\x=\pm\frac{1}{2}+k2\pi\end{array}\right.\)với k\(\in\)Z
pt có 2 nghiệm như trên
pi<x<3/2pi
=>cosx<0
pi<x<3/2pi
=>pi/2<1/2x<3/4pi
=>cos(x/2)<0
1+tan^2x=1/cos^2x
=>1/cos^2x=1+8=9
=>cosx=-1/3
\(cosx=2\cdot cos^2\left(\dfrac{x}{2}\right)-1\)
=>\(2\cdot cos^2\left(\dfrac{x}{2}\right)=\dfrac{2}{3}\)
=>\(cos^2\left(\dfrac{x}{2}\right)=\dfrac{1}{3}\)
=>cos(x/2)=1/căn 3
Lời giải:
$\sin (2x+1)=\frac{-1}{2}$
$\Rightarrow 2x+1=\frac{-\pi}{6}+2k\pi$ hoặc $2x+1=\frac{7}{6}\pi +2k\pi$ với $k$ nguyên
Với $2x+1=\frac{-\pi}{6}+2k\pi$
Do $x\in (0;\pi)$ nên $k=1$
$x=\frac{11}{12}\pi -\frac{1}{2}$
Với $2x+1=\frac{7\pi}{6}+2k\pi$
Do $x\in (0;\pi)$ nên $k=0$
$\Rightarrow x=\frac{7}{12}\pi -\frac{1}{2}$
\(\Leftrightarrow2cos^2x-1+9cosx+4=0\)
\(\Leftrightarrow2cos^2x+9cosx+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{-9-\sqrt{57}}{4}\left(l\right)\\cosx=\dfrac{-9+\sqrt{57}}{4}\left(nh\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\alpha+k2\pi\)