câu 1:xét sinx=o

xét sinx khác 0

chia phương trình cho cos³x

ta được 1 phương trình mới:

4+3tanx-\(\frac{1}{sin^2x}\)-tan³x=0

<=>4+3tanx-(1+cot²x)-tan³x=0

<=>4+3tanx-1-\(\frac{1}{tan^2x}\)-tan³x=o

nhân cho tan²x ta được 1 phương trình bậc 5 với tanx

Đây nè bn:

\(A=\frac{\sqrt{3}sinx.\left(cosx.cos\frac{\pi}{6}-sinx.sin\frac{\pi}{6}\right)+cosx\left(sin\frac{\pi}{3}cosx-cos\frac{\pi}{6}.sinx\right)}{sin\left(2x+\frac{\pi}{3}\right)}\)

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

\(A=\frac{sinx.cosx+\frac{\sqrt{3}}{2}\left(cos^2x-sin^2x\right)}{sin\left(2x+\frac{\pi}{3}\right)}\)

\(A=\frac{\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x}{sin\left(2x+\frac{\pi}{3}\right)}=\frac{sin2x.cos\frac{\pi}{3}+cos2x.sin\frac{\pi}{3}}{sin\left(2x+\frac{\pi}{3}\right)}\)

\(A=\frac{sin\left(2x+\frac{\pi}{3}\right)}{sin\left(2x+\frac{\pi}{3}\right)}=1\)

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

c/

ĐKXĐ: \(sin4x\ne0\Leftrightarrow x\ne\frac{k\pi}{4}\)

\(\Leftrightarrow\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=2\)

\(\Leftrightarrow4sinx.cos2x+2cos2x=2\)

\(\Leftrightarrow cos2x\left(2sinx+1\right)=1\)

\(\Leftrightarrow\left(1-2sin^2x\right)\left(2sinx+1\right)=1\)

\(\Leftrightarrow4sin^3x+2sin^2x-2sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+sinx-1\right)=0\)

\(\Leftrightarrow2sin^2x+sinx-1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

b/

Nhận thấy \(sinx=0\) không phải nghiệm, pt tương đương:

\(sinx.cosx.cos2x.cos4x.cos8x=\frac{1}{16}sinx\)

\(\Leftrightarrow8sin2x.cos2x.cos4x.cos8x=sinx\)

\(\Leftrightarrow4sin4x.cos4x.cos8x=sinx\)

\(\Leftrightarrow2sin8x.cos8x=sinx\)

\(\Leftrightarrow sin16x=sinx\)

\(\Rightarrow\left[{}\begin{matrix}16x=x+k2\pi\\16x=\pi-x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k2\pi}{15}\\x=\frac{\pi}{17}+\frac{k2\pi}{17}\end{matrix}\right.\)

2.1

a.

\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)

b.

\(cosx-\sqrt{3}sinx=1\)

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

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

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