\(cot^2x-cotx=0\)
Giải phương trình:
\(2\left(Tan^2x-Cot^2x\right)-5\left(Tanx+Cotx\right)+6=0\)
Giải pt sau:
\(2\left(Tan^2x-Cot^2x\right)-5\left(Tanx+Cotx\right)+6=0\)
Cho \(tanx-cotx=3\). Tính giá trị của biểu thức : \(A=tan^2x+cot^2x;B=tanx+cotx;C=tan^4x-cot^4x\)
\(\left(tanx-cotx\right)^2=9\Rightarrow tan^2x-2.tanx.cotx+cot^2x=9\)
\(\Rightarrow tan^2x+cot^2x=11\)
\(\left(tanx+cotx\right)^2=tan^2x+cot^2x+2.tanx.cotx=11+2=13\)
\(\Rightarrow tanx+cotx=\pm\sqrt{13}\)
\(tan^4x-cot^4x=\left(tan^2x+cot^2x\right)\left(tan^2x-cot^2x\right)\)
\(=11\left(tanx+cotx\right)\left(tanx-cotx\right)=\pm33\sqrt{13}\)
Đơn giản biểu thức : O = \(\frac{cot^2x-cos^2x}{cot^2x}+\frac{sinx.cosx}{cotx}\)
\(O=\frac{cot^2x}{cot^2x}-\frac{cos^2x}{cot^2x}+sinx.cosx.tanx\)
\(=1-cos^2x.\frac{sin^2x}{cos^2x}+sinx.cosx.\frac{sinx}{cosx}\)
\(=1-sin^2x+sin^2x=1\)
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+tan^2x+tan^3x+cotx+cot^2x+cot^3x=6
Giải hộ mk cái !!!
1. tan(2x + 10°) + cotx =0
2. tan(2x + 45°) . tan(180° - x/2) =1
3. cot(3x + 45°) - tan20° =0
tan(2x+10o)+cot(x)=0
<=> tan(2x+10o)+tan(90o-x)=0
<=>tan(x+100o)*[1-tan(2x-10o)*tan(90o-x)]=0
*tan(x+100o)=0 => x=....
*1-tan(2x-10o)*tan(90o-x)=0
<=> tan(2x-10o)=tanx <=> x=....
\(\int\dfrac{cotx}{sin^2x}dx\) = ?
A. \(-\dfrac{cot^2x}{2}+c\)
B. \(\dfrac{cot^2x}{2}+c\)
C. \(\dfrac{-tan^2x}{2}+c\)
D. \(\dfrac{tan^2x}{2}+c\)
Giải PT
a) sin2 x + 2sinx - 3 = 0
b) 2cos x + cos 2x = 0
c) tanx + cotx + 2 = 0
d) sinx + cos2x + 1 = 0
e) tan x + cot 2x = 0
a) TH1: sinx = 1
--> x = pi/2 + k2pi (k nguyên)
TH2: sinx = -3 (loại)
b) 2cosx + cos2x = 0
<=> 2cosx + 2cos^2(x) - 1 = 0
TH1: cosx = (-1 + sqrt(3))/2
TH2: cosx = (-1 - sqrt(3))/2 (loại)
c) ĐKXĐ: x # kpi
Pt <=> tanx + 1/tanx + 2 = 0
--> tanx = -1
--> x = -pi/4 + kpi (k nguyên)