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\(\dfrac{1}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{1+sinx-cos^2x}{cosx\left(1+sinx\right)}\)
\(=\dfrac{\left(1+sinx\right)-\left(1+sinx\right)\left(1-sinx\right)}{cosx\left(1+sinx\right)}\)
\(=\dfrac{\left(1+sinx\right)\left(1-1+sinx\right)}{\left(1+sinx\right)\cdot cosx}=\dfrac{sinx}{cosx}=tanx\)
=>ĐPCM
1.
\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
2.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
1.
ĐK: \(x\ne\dfrac{k\pi}{2}\)
\(cotx-tanx=sinx+cosx\)
\(\Leftrightarrow\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=sinx+cosx\)
\(\Leftrightarrow\dfrac{cos^2x-sin^2x}{sinx.cosx}=sinx+cosx\)
\(\Leftrightarrow\left(\dfrac{cosx-sinx}{sinx.cosx}-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(1\right)\\cosx-sinx=sinx.cosx\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow t=\dfrac{1-t^2}{2}\left(t=cosx-sinx,\left|t\right|\le2\right)\)
\(\Leftrightarrow t^2+2t-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1+\sqrt{2}\\t=-1-\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow cosx-sinx=-1+\sqrt{2}\)
\(\Leftrightarrow-\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=-1+\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}-1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)
1.
\(8sinx=\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\)
\(\Leftrightarrow4sinx=\dfrac{\sqrt{3}}{2cosx}+\dfrac{1}{2sinx}\)
\(\Leftrightarrow4sinx=\dfrac{\sqrt{3}sinx+cosx}{sin2x}\)
\(\Leftrightarrow4sinx.sin2x=\sqrt{3}sinx+cosx\)
\(\Leftrightarrow2cosx-2cos3x=\sqrt{3}sinx+cosx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=2cos3x\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos3x\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=\pm3x+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}-k\pi\\x=-\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)
2.
ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(sinx+\sqrt{3}cosx=\dfrac{1}{cosx}\)
\(\Leftrightarrow2sinx.cosx+2\sqrt{3}cos^2x-\sqrt{3}=2-\sqrt{3}\)
\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=1-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{2-\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k2\pi\\2x+\dfrac{\pi}{3}=\pi-arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+\dfrac{1}{2}arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k\pi\\x=\dfrac{\pi}{3}-\dfrac{1}{2}arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k\pi\end{matrix}\right.\)
Đk: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+m2\pi\\x\ne\dfrac{\pi}{4}+n\pi\end{matrix}\right.\left(m,n\in Z\right)\)
PT \(\Leftrightarrow1=2\sqrt{2}sinx.cosx\left(sinx-cosx\right)+2cos^2x\)
\(\Leftrightarrow\sqrt{2}.2sinx.cosx\left(sinx-cosx\right)+\left(2cos^2x-1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin2x\left(sinx-cosx\right)+\left(cosx-sinx\right)\left(cosx+sinx\right)=0\)
\(\Leftrightarrow\sqrt{2}sin2x=sinx+cosx\)
\(\Leftrightarrow\sqrt{2}sin2x=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{4}+k2\pi\\2x=\pi-x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+k\dfrac{2\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)
Ta có \(sinx-cosx=\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\)
a, Do \(0< x< \dfrac{\pi}{4}\Rightarrow-\dfrac{\pi}{4}< x-\dfrac{\pi}{4}< 0\)
⇒ \(\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\) < 0
⇒ sinx - cosx < 0
=> sinx < cosx
b, Do \(\dfrac{\pi}{4}< x< \dfrac{\pi}{2}\Rightarrow0< x-\dfrac{\pi}{4}< \dfrac{\pi}{4}\)
⇒ \(\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\) > 0
⇒ sinx - cosx > 0
=> sinx > cosx
3/2pi<x<2pi
=>sin x<0; cosx>0
sin x+cosx=-1/2
=>(sinx+cosx)^2=1/4
=>1+2*sinx*cosx=1/4
=>2*sin x*cosx=-3/4
=>sinx*cosx=-3/8
mà sin x+cosx=-1/2
nên \(sinx=\dfrac{-1-\sqrt{7}}{4};cosx=\dfrac{-1+\sqrt{7}}{4}\)
a: ĐKXĐ: \(cosx-1\ne0\)
=>\(cosx\ne1\)
=>\(x\ne k2\Omega\)
b: ĐKXĐ: sin x-1>=0
=>sin x>=1
mà \(-1< =sinx< =1\)
nên sin x=1
=>\(x=\dfrac{\Omega}{2}+k2\Omega\)
c:
-1<=sin x<=1
=>-1+1<=sin x+1<=1+1
=>0<=sin x+1<=2
ĐKXĐ: \(\dfrac{1+sinx}{1-cosx}>=0\)
mà \(1+sinx>=0\)(cmt)
nên \(1-cosx>0\)
=>\(cosx< 1\)
mà -1<=cosx<=1
nên \(cosx\ne1\)
=>\(x\ne k2\Omega\)
\(\dfrac{2}{sinx}-\dfrac{sinx}{1+cosx}\)
\(=\dfrac{2+2cosx-sin^2x}{sinx\left(1+cosx\right)}=\dfrac{2\left(1+cosx\right)-\left(1-cos^2x\right)}{sinx\left(1+cosx\right)}\)
\(=\dfrac{\left(1+cosx\right)\left(2-1+cosx\right)}{sinx\left(1+cosx\right)}=\dfrac{cosx+1}{sinx}\)