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cotx - cot2x = tanx + 1 (1)
Điều kiện: sinx ≠ 0 và cosx ≠ 0. Khi đó:
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:
\(x=-\dfrac{\pi}{4}+k\pi;x=\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi;x=\dfrac{5\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)+k2\pi\)
ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\)
\(\frac{1}{\frac{sinx}{cosx}+\frac{cos2x}{sin2x}}=\frac{\sqrt{2}\left(cosx-sinx\right)}{\frac{cosx}{sinx}-1}\)
\(\Leftrightarrow\frac{sin2x.cosx}{cos2x.cosx+sin2x.sinx}=\frac{\sqrt{2}sinx\left(cosx-sinx\right)}{cosx-sinx}\)
\(\Leftrightarrow\frac{sin2x.cosx}{cosx}=\sqrt{2}sinx\)
\(\Leftrightarrow2sinx.cosx=\sqrt{2}sinx\)
\(\Leftrightarrow cosx=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\left(l\right)\\x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
Vậy \(x=-\frac{\pi}{4}+k2\pi\)
Đặt \(x+\frac{\pi}{4}=t\Rightarrow x=t-\frac{\pi}{4}\)
Pt trở thành:
\(sin^3t=\sqrt{2}sin\left(t-\frac{\pi}{4}\right)\)
\(\Leftrightarrow sin^3t=sint-cost\)
\(\Leftrightarrow sint-sin^3t-cost=0\)
\(\Leftrightarrow sint\left(1-sin^2t\right)-cost=0\)
\(\Leftrightarrow sint.cos^2t-cost=0\)
\(\Leftrightarrow cost\left(sint.cost-1\right)=0\)
\(\Leftrightarrow cost\left(\frac{1}{2}sin2t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=2>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cos\left(x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
c/
ĐKXĐ: ...
Chia 2 vế cho \(cos^2x\) ta được:
\(\left(1+tanx\right)tan^2x=3tanx\left(1-tanx\right)+3\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^3x+tan^2x=3tanx-3tan^2x+3+3tan^2x\)
\(\Leftrightarrow tan^3x+tan^2x-3tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow\left[{}\begin{matrix}\pi\left(x+1\right)=\frac{\pi}{3}+k2\pi\\\pi\left(x+1\right)=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\frac{1}{3}+2k\\x+1=-\frac{1}{3}+2k\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}+2k\\x=-\frac{4}{3}+2k\end{matrix}\right.\)
a.
ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow tanx=\frac{1}{cot2x}\)
\(\Leftrightarrow tanx=tan2x\)
\(\Leftrightarrow2x=x+k\pi\)
\(\Leftrightarrow x=k\pi\) (ktm)
Vậy pt vô nghiệm
b. ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cos2x}{sin2x}=sinx\)
\(\Leftrightarrow\frac{cos2x.cosx+sin2x.sinx}{cosx.sin2x}=sinx\)
\(\Leftrightarrow\frac{cosx}{cosx.sin2x}=sinx\)
\(\Leftrightarrow\frac{1}{sin2x}=sinx\Leftrightarrow sin2x.sinx=1\)
Do \(\left\{{}\begin{matrix}sin2x\le1\\sinx\le1\end{matrix}\right.\) nên \(sin2x.sinx\le1\)
Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sin2x=1\\sinx=1\end{matrix}\right.\) (vô nghiệm)
Vậy pt đã cho vô nghiệm
1: tan x=3 nên sin x/cosx=3
=>sin x=3*cosx
\(B=\dfrac{2\cdot sinx-3cosx}{sinx+cosx}=\dfrac{2\cdot3\cdot cosx-3cosx}{3cosx+cosx}\)
\(=\dfrac{2\cdot3-3}{3+1}=\dfrac{3}{4}\)
2: tan x=-1 nên sin x/cosx=-1
=>sinx=-cosx
\(I=\dfrac{4\cdot\left(-cosx\right)^3+\left(cosx\right)^3}{-cosx+3\cdot cosx}=\dfrac{-3\cdot cos^3x}{2cosx}=-\dfrac{3}{2}\cdot cos^2x\)
\(1+tan^2x=\dfrac{1}{cos^2x}\)
=>\(\dfrac{1}{cos^2x}=1+1=2\)
=>\(cos^2x=\dfrac{1}{2}\)
=>I=-3/2*1/2=-3/4
b/ ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\left(1-\frac{sinx}{cosx}\right)\left(1+sinx\right)=1+\frac{sinx}{cosx}\)
\(\Leftrightarrow\left(cosx-sinx\right)\left(1+sinx\right)=sinx+cosx\)
\(\Leftrightarrow cosx+sinx.cosx-sinx-sin^2x=sinx+cosx\)
\(\Leftrightarrow sin^2x+2sinx-sinx.cosx=0\)
\(\Leftrightarrow sinx\left(sinx-cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sinx-cosx=-2\left(1\right)\end{matrix}\right.\)
Xét \(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=-2\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\) (vô nghiệm)
a/ ĐKXĐ: \(sin4x\ne0\)
\(\frac{sinx}{cosx}+\frac{cos2x}{sin2x}=\frac{2cos4x}{sin4x}\)
\(\Leftrightarrow2sin^2x.cos2x+2cos^22x=2cos4x\)
\(\Leftrightarrow\left(1-cos2x\right)cos2x+2cos^22x=4cos^22x-2\)
\(\Leftrightarrow3cos^22x-cos2x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\left(l\right)\\cos2x=-\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow2x=\pm arccos\left(-\frac{2}{3}\right)+k2\pi\)
\(\Leftrightarrow x=\pm\frac{1}{2}arccos\left(-\frac{2}{3}\right)+k\pi\)