a/

\(\Leftrightarrow cos^3x-sin^3x=cosx+sinx\)

- Với \(cosx=0\Rightarrow sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\) là 1 nghiệm

- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow1-tan^3x=\frac{1}{cos^2x}+tanx.\frac{1}{cos^2x}\)

\(\Leftrightarrow1-tan^3x=1+tan^2x+tanx\left(1+tan^2x\right)\)

\(\Leftrightarrow2tan^3x+tan^2x+tanx=0\)

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

\(\Leftrightarrow tanx=0\Rightarrow x=k\pi\)

b/

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}=1+2sinx\)

\(\Leftrightarrow\frac{cosx-sinx}{cosx+sinx}=1+2sinx\)

\(\Leftrightarrow cosx-sinx=\left(1+2sinx\right)\left(cosx+sinx\right)\)

\(\Leftrightarrow sinx+sinx.cosx+sin^2x=0\)

\(\Leftrightarrow sinx\left(sinx+cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sinx+cosx=-1\left(1\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\left(l\right)\\x=\pi+k2\pi\end{matrix}\right.\)

c/

ĐKXĐ: ...

Chia 2 vế cho \(cos^2x\) ta được:

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)

a.

a/

\(\Leftrightarrow2cos2x.cosx+\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right).cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos^22x=0\)

\(\Leftrightarrow cos2x\left(2cosx+cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\left(1\right)\\2cosx+cos2x=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x=\frac{\pi}{2}+k\pi\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

\(\left(2\right)\Leftrightarrow2cosx+2cos^2x-1=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}-1}{2}\\cosx=\frac{-\sqrt{3}-1}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm arccos\left(\frac{\sqrt{3}-1}{2}\right)+k2\pi\)

b/

\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cosx+1-cos^2x+2cos^2x-1=\frac{1}{2}\)

\(\Leftrightarrow cos^2x+\frac{1}{2}cosx=0\)

\(\Leftrightarrow cosx\left(cosx+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\left(\frac{sinx}{cosx}+\frac{cosx}{sinx}\right)^2+\frac{3}{sin2x}-7=0\)

\(\Leftrightarrow\left(\frac{sin^2x+cos^2x}{sinx.cosx}\right)^2+\frac{3}{sin2x}-7=0\)

\(\Leftrightarrow\left(\frac{2}{sin2x}\right)^2+\frac{3}{sin2x}-7=0\)

Đặt \(\frac{1}{sin2x}=a\Rightarrow4a^2+3a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{7}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{4}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k2\pi\\2x=arcsin\left(-\frac{4}{7}\right)+k2\pi\\2x=\pi-arcsin\left(-\frac{4}{7}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{4}{7}\right)+k\pi\end{matrix}\right.\)

a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)

\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)

\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)

\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)

b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx+\sqrt{3}\cdot cosx=0\)

\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)

\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)

(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)

\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)

\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)

\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)

\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)

\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)

b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)

=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)

d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)

\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)

=\(\frac{1}{cosx.sinx}=VP\)

e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)

c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)

=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)

\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)

Đây nha bạn

đáp án không giống lắm 

2: \(\left(sinx+cosx\right)^2=1+2\cdot sinx\cdot cosx=1+2\cdot\dfrac{\sqrt{3}}{4}=1+\dfrac{\sqrt{3}}{2}=\dfrac{2+\sqrt{3}}{2}\)

=>\(sinx+cosx=\dfrac{\sqrt{3}+1}{2}\)

mà sin x*cosx=căn 3/4

nên sinx,cosx là các nghiệm của phương trình là:

\(a^2-\dfrac{\sqrt{3}+1}{2}\cdot a+\dfrac{\sqrt{3}}{4}=0\)

=>\(\left[{}\begin{matrix}a=\dfrac{\sqrt{3}}{2}\\a=\dfrac{1}{2}\end{matrix}\right.\)

Ta sẽ có hai trường hợp:

TH1: sin x=căn 3/2; cosx=1/2

tan x=sinx/cosx=căn 3

cot x=1/căn 3

TH2: sin x=1/2; cosx=căn 3/2

tan x=sin x/cosx=1/căn 3

cot x=1:1/căn 3=căn 3