Lời giải:

a. 

\(A=\frac{3}{2}-2(\frac{\cos x}{\sin x})^2=\frac{3}{2}-2.(\frac{1}{\tan x})^2=\frac{3}{2}-\frac{1}{2}(\frac{-3}{2})^2=-3\)

b.

\(A=\frac{1}{2}(\frac{\sin x}{\cos x})^2-\frac{5}{2}=2(\frac{1}{\cot x})^2-\frac{5}{2}=2(\frac{5}{3})^2-\frac{5}{2}=\frac{55}{18}\)

a, \(A=\dfrac{3sin^2\left(x\right)-cos^2\left(x\right)}{2sin^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\dfrac{cos^2\left(x\right)}{sin^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\cdot\dfrac{1}{tan^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\cdot\left(-\dfrac{3}{2}\right)^2=-3\)

b, \(A=\dfrac{sin^2\left(x\right)-5cos^2\left(x\right)}{2cos^2\left(x\right)}=\dfrac{1}{2}\dfrac{sin^2\left(x\right)}{cos^2\left(x\right)}-\dfrac{5}{2}=\dfrac{1}{2}\cdot\dfrac{1}{cot^2\left(x\right)}-\dfrac{5}{2}=\dfrac{1}{2}\cdot\left(\dfrac{5}{3}\right)^2-\dfrac{5}{2}=\dfrac{55}{18}\)

:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))

Giả sử các biểu thức đều có nghĩa

\(A=2\left(\left(sin^2x\right)^3+\left(cos^2x\right)^3\right)-3\left(sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x\right)\)

\(A=2\left(sin^2x+cos^2x\right)\left(\left(sin^2x+cos^2x\right)^2-3sin^2xcos^2x\right)-3\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)\)

\(A=2\left(1-3sin^2xcos^2x\right)-3\left(1-2sin^2xcos^2x\right)\)

\(A=2-6sin^2xcos^2x-3+6sin^2xcos^2x=-1\)

b/ \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{\dfrac{1}{cotx}-1}\)

\(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2cotx}{1-cotx}=\dfrac{1+cotx-2cotx}{1-cotx}=\dfrac{1-cotx}{1-cotx}=1\)

c/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)

\(C=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)

\(C=cos^2x-sin^2x+cos^2x+3sin^2x\)

\(C=2cos^2x+2sin^2x=2\left(cos^2x+sin^2x\right)=2\)

a/

\(\Leftrightarrow\left(sin^2\frac{x}{3}+cos^2\frac{x}{3}\right)^2-2sin^2\frac{x}{3}.cos^2\frac{x}{3}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}sin^2\frac{2x}{3}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{4}\left(1-cos\frac{4x}{3}\right)=\frac{5}{8}\)

\(\Leftrightarrow cos\frac{4x}{3}=-\frac{1}{2}\)

\(\Leftrightarrow\frac{4x}{3}=\pm\frac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\frac{\pi}{2}+\frac{k3\pi}{2}\)

b/

\(\Leftrightarrow4\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow4-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\)

\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-1\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x+\frac{1}{2}cos4x=-\frac{1}{2}\)

\(\Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=-\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=-\frac{\pi}{6}+k2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)

c/

\(\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)

\(\Leftrightarrow-cos^32x+5cos^22x-7cos2x+3=0\)

\(\Leftrightarrow\left(3-cos2x\right)\left(cos2x-1\right)^2=0\)

\(\Leftrightarrow cos2x=1\)

\(\Leftrightarrow x=k\pi\)

d/

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos4x\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=cos4x\)

\(\Leftrightarrow1-\frac{3}{8}\left(1-cos4x\right)=cos4x\)

\(\Leftrightarrow cos4x=1\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

1: cot x=-6 nên cosx/sinx=-6

=>cosx=-6*sinx

\(F=\dfrac{sinx-3\cdot cosx}{cosx+2\cdot sinx}=\dfrac{sinx+18\cdot sinx}{-6\cdot sinx+2\cdot sinx}=\dfrac{20}{-4}=-5\)

2: cotx=1

=>cosx/sinx=1

=>cosx=sinx

\(I=\dfrac{sin^3x-4\cdot sin^3x}{sinx+3sinx}=\dfrac{5\cdot sin^3x}{4\cdot sinx}=\dfrac{5}{4}\cdot sin^2x\)

\(1+cot^2x=\dfrac{1}{sin^2x}\)

=>\(\dfrac{1}{sin^2x}=1+1=2\)

=>sin^2=1/2

=>\(I=\dfrac{5}{4}\cdot\dfrac{1}{2}=\dfrac{5}{8}\)

3: cotx=3

=>cosx/sinx=3

=>cosx=3*sinx

1+cot^2x=1/sin^2x

=>\(\dfrac{1}{sin^2x}=1+9=10\)

=>\(sin^2x=\dfrac{1}{10}\)

\(I=\dfrac{2\cdot sin^3x+cos^3x}{4\cdot sinx-6\cdot cosx}\)

\(=\dfrac{2\cdot sin^3x+\left(3\cdot sinx\right)^3}{4\cdot sinx-6\cdot\left(3\cdot sinx\right)}=\dfrac{2\cdot sin^3x+27\cdot sin^3x}{4\cdot sinx-18\cdot sinx}\)

\(=\dfrac{29}{-14}\cdot sin^2x=\dfrac{-29}{14}\cdot\dfrac{1}{10}=-\dfrac{29}{140}\)

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)

cotx=2

=>cosx=2*sin x

\(1+cot^2x=\dfrac{1}{sin^2x}\)

=>\(\dfrac{1}{sin^2x}=1+4=5\)

=>\(sin^2x=\dfrac{1}{5}\)

\(B=\dfrac{sin^2x-2\cdot sinx\cdot2\cdot sinx-1}{5\cdot4sin^2x+sin^2x-3}=\dfrac{-3sin^2x-1}{21sin^2x-3}\)

\(=\dfrac{-\dfrac{3}{5}-1}{\dfrac{21}{5}-3}=-\dfrac{8}{5}:\dfrac{6}{5}=-\dfrac{4}{3}\)

\(cotx=2\Rightarrow tanx=\dfrac{1}{2}\)

\(B=\dfrac{sin^2x-2sinx.cosx-1}{5cos^2x+sin^2x-3}\)

\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-\dfrac{1}{cos^2x}}{5+tan^2x-\dfrac{3}{cos^2x}}\)

\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-1-tan^2x}{5+tan^2x-3-3tan^2x}\)

\(\Leftrightarrow B=\dfrac{-2tanx-1}{2-2tan^2x}\)

\(\Leftrightarrow B=\dfrac{-2.\dfrac{1}{2}-1}{2-2.\dfrac{1}{4}}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)

\(A=\frac{\frac{sin^2x}{cos^2x}+\frac{sinx.cosx}{cos^2x}+\frac{5}{cos^2x}}{\frac{3sin^2x}{cos^2x}-\frac{2cos^2x}{cos^2x}}=\frac{tan^2x+tanx+5\left(1+tan^2x\right)}{3tan^2x-2}\)

\(=\frac{\left(-3\right)^2-3+5\left[1+\left(-3\right)^2\right]}{3.\left(-3\right)^2-2}=...\)

d.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^4x\)

\(tan^4x-3tan^2x-4tanx-3=0\)

\(\Leftrightarrow\left(tan^2x+tanx+1\right)\left(tan^2x-tanx-3\right)=0\)

\(\Leftrightarrow tan^2x-tanx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1-\sqrt{13}}{2}\\tanx=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(\frac{1-\sqrt{13}}{2}\right)+k\pi\\x=arctan\left(\frac{1+\sqrt{13}}{2}\right)+k\pi\end{matrix}\right.\)

mọi người giúp hộ mình nhanh với

a.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(2tan^3x+4=3tanx\left(1+tan^2x\right)\)

\(\Leftrightarrow2tan^3x+4=3tanx+3tan^3x\)

\(\Leftrightarrow tan^3x+3tanx-4=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x+tanx+4\right)=0\)

\(\Leftrightarrow tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\)