GPT
\(tanx+cotx+14=cot^22x\)
Cho \(tanx-cotx=3\). Tính giá trị của biểu thức : \(A=tan^2x+cot^2x;B=tanx+cotx;C=tan^4x-cot^4x\)
\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x-2.tanx.cotx+cot^2x=9\)
\(\Rightarrow tan^2x+cot^2x=11\)
\(\left(tanx+cotx\right)^2=tan^2x+cot^2x+2.tanx.cotx=11+2=13\)
\(\Rightarrow tanx+cotx=\pm\sqrt{13}\)
\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)
\(=11\left(tanx+cotx\right)\left(tanx-cotx\right)=\pm33\sqrt{13}\)
giải các pt
a) \(cos2x+cosx+1=0\)
b) \(tanx+cotx=2\)
c) \(tan^2x+\left(\sqrt{3}-1\right)tanx-\sqrt{3}=0\)
d) \(cot^22x+\frac{3}{tan2x}+2=0\)
a/
\(\Leftrightarrow2cos^2x-1+cosx+1=0\)
\(\Leftrightarrow cosx\left(2cosx+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(\Leftrightarrow tanx+\frac{1}{tanx}=2\)
\(\Leftrightarrow tan^2x+1=2tanx\)
\(\Leftrightarrow tan^2x-2tanx+1=0\)
\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)
c/
\(a+b+c=1+\sqrt{3}-1-\sqrt{3}=0\)
\(\Rightarrow\) Pt có 2 nghiệm: \(\left[{}\begin{matrix}tanx=1\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
d/ ĐKXĐ: ...
\(\Leftrightarrow cot^22x+3.cot2x+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cot2x=-1\\cot2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{4}+k\pi\\2x=arccot\left(-2\right)+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{8}+\frac{k\pi}{2}\\x=\frac{1}{2}arccot\left(-2\right)+\frac{k\pi}{2}\end{matrix}\right.\)
tanx +cotx +7 = cot^2 2x
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}+7=\dfrac{cos^22x}{sin^22x}\)
\(\Leftrightarrow\dfrac{sin^2x+cos^2x}{sinx.cosx}+7=\dfrac{1-sin^22x}{sin^22x}\)
\(\Leftrightarrow\dfrac{2}{sin2x}+7=\dfrac{1}{sin^22x}-1\)
\(\Leftrightarrow\dfrac{1}{sin^22x}-\dfrac{2}{sin2x}-8=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{sin2x}=4\\\dfrac{1}{sin2x}=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=\dfrac{1}{4}\\sin2x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}arcsin\left(\dfrac{1}{4}\right)+k\pi\\x=\dfrac{\pi}{2}-\dfrac{1}{2}arcsin\left(\dfrac{1}{4}\right)+k\pi\\x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{matrix}\right.\)
(tanx+7)tanx+(cotx+7)cotx=-14
ĐKXĐ: \(x\notin\left\{\dfrac{\Omega}{2}+k\Omega;\Omega+k\Omega\right\}\)
(tanx+7)*tanx+(cotx+7)*cotx=-14
=>\(tan^2x+cot^2x+7\left(tanx+cotx\right)=-14\)
=>\(\left(tanx+cotx\right)^2-2\cdot cotx\cdot tanx+7\left(tanx+cotx\right)+14=0\)
=>\(\left(tanx+cotx\right)^2+7\left(tanx+cotx\right)+12=0\)
=>\(\left(tanx+\dfrac{1}{tanx}+3\right)\left(tanx+\dfrac{1}{tanx}+4\right)=0\)
=>\(\dfrac{tan^2x+3tanx+1}{tanx}\cdot\dfrac{tan^2x+4tanx+1}{tanx}=0\)
=>\(\left[{}\begin{matrix}tan^2x+3tanx+1=0\\tan^2x+4tanx+1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}tanx=\dfrac{-3+\sqrt{5}}{2}\\tanx=\dfrac{-3-\sqrt{5}}{2}\\tanx=-2+\sqrt{3}\\tanx=-2-\sqrt{3}\end{matrix}\right.\)
=>\(x\in\left\{arctan\left(\dfrac{-3+\sqrt{5}}{2}\right)+k\Omega;arctan\left(\dfrac{-3-\sqrt{5}}{2}\right)+k\Pi;arctan\left(-2+\sqrt{3}\right)+k\Omega;arctan\left(-2-\sqrt{3}\right)+k\Omega\right\}\)
tanx+tan^2x+tan^3x+cotx+cot^2x+cot^3x=6
Giải hộ mk cái !!!
Tan2x+cot2x+2(tanx+cotx)=6
Giải phương trình:
\(2\left(Tan^2x-Cot^2x\right)-5\left(Tanx+Cotx\right)+6=0\)
Giải pt sau:
\(2\left(Tan^2x-Cot^2x\right)-5\left(Tanx+Cotx\right)+6=0\)
\(GPT:2\sqrt{3}.cotx-\frac{1}{sinx}=1+\frac{\sqrt{3}cotx}{sinx}-cot^2x\)