\(\Leftrightarrow\left[{}\begin{matrix}5x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+k2\pi\\5x+\dfrac{\pi}{6}=\pi-\left(x-\dfrac{\pi}{3}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-\dfrac{\pi}{2}+k2\pi\\6x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+k\dfrac{\pi}{2}\\x=\dfrac{7\pi}{36}+k\dfrac{\pi}{3}\end{matrix}\right.\left(k\in Z\right)\)

\(2sinx+2\sqrt{3}cosx-\sqrt{3}sin2x+cos2x=\sqrt{3}cosx+cos2x-2sinx+2\)

\(\Leftrightarrow4sinx+\sqrt{3}cosx-2\sqrt{3}sinx.cosx-2=0\)

\(\Leftrightarrow-2sinx\left(\sqrt{3}cosx-2\right)+\sqrt{3}cosx-2=0\)

\(\Leftrightarrow\left(1-2sinx\right)\left(\sqrt{3}cosx-2\right)=0\)

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

\(\Leftrightarrow...\)

\(\sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{2}\Leftrightarrow x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{3\pi}{4}+k2\pi\left(k\in Z\right)\)

`sin(x- (pi)/4) = (pi)/2`

`<=> x - (pi)/4 = (pi)/2 + k2(pi)`

`<=> x = (3(pi))/4 + k2(pi)`.

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

\(\Leftrightarrow x=\pi+k2\pi\left(k\in Z\right)\)

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

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\)

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

\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\left(k\in Z\right)\)

`sin (x-\pi/3)=1`

`<=>x-\pi/3=\pi/2+k2\pi` , `k in ZZ`

`<=>x=[5\pi]/6 +k2\pi` , `k in ZZ`

Vậy ptr có nghiệm `x=[5\pi]/6 +k2\pi` , `k in ZZ`

\(sin\left(x\right)+\left[sin\left(x+\dfrac{2\pi}{5}\right)-sin\left(x+\dfrac{\pi}{5}\right)\right]+\left[sin\left(x+\dfrac{4\pi}{5}\right)-sin\left(x+\dfrac{3\pi}{5}\right)\right]\)

\(=sin\left(x\right)+2cos\left(x+\dfrac{3\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)+2cos\left(x+\dfrac{7\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)\)

\(=sin\left(x\right)+2sin\left(\dfrac{\pi}{10}\right)\left[cos\left(x+\dfrac{3\pi}{10}\right)+cos\left(x+\dfrac{7\pi}{10}\right)\right]\)

\(=sin\left(x\right)+4sin\left(\dfrac{\pi}{10}\right)cos\left(\dfrac{\pi}{5}\right)cos\left(x+\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)+cos\left(x+\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)+cos\left(x\right)cos\left(\dfrac{\pi}{2}\right)-sin\left(x\right)sin\left(\dfrac{\pi}{2}\right)\)

\(=sin\left(x\right)-sin\left(x\right)\)

\(=0\)

Ta có: \(\sin x+\sin\left(x+\frac45\pi\right)\)

\(=2\cdot\sin\left(\frac{x+x+\frac45\pi}{2}\right)\cdot cos\left(\frac{x+\frac45\pi-x}{2}\right)=2\cdot\sin\left(x+\frac25\pi\right)\cdot cos\left(\frac25\pi\right)\)

Ta có: \(\sin\left(x+\frac{\pi}{5}\right)+\sin\left(x+\frac35\pi\right)\)

\(=2\cdot\sin\left(\frac{x+\frac{\pi}{5}+x+\frac35\pi}{2}\right)\cdot cos\left(\frac{x+\frac35\pi-x-\frac{\pi}{5}}{2}\right)\)

\(=2\cdot\sin\left(x+\frac25\pi\right)\cdot cos\left(\frac{\pi}{5}\right)\)

Ta có: \(Q=\sin x-\sin\left(x+\frac{\pi}{5}\right)+\sin\left(x+\frac25\pi\right)-\sin\left(x+\frac35\pi\right)+\sin\left(x+\frac45\pi\right)\)

\(=2\cdot\sin\left(x+\frac25\pi\right)\cdot cos\left(\frac25\pi\right)-2\cdot\sin\left(x+\frac25\pi\right)\cdot cos\left(\frac{\pi}{5}\right)+\sin\left(x+\frac25\pi\right)\)

\(=\sin\left(x+\frac25\pi\right)\left\lbrack2\cdot cos\left(\frac25\pi\right)-2\cdot cos\left(\frac{\pi}{5}\right)+1\right\rbrack\)

\(=\sin\left(x+\frac25\pi\right)\cdot\left\lbrack2\cdot\left(2\cdot cos^2\left(\frac{\pi}{5}\right)-1\right)-2\cdot cos\left(\frac{\pi}{5}\right)+1\right\rbrack\)

\(=\sin\left(x+\frac25\pi\right)\cdot\left\lbrack4\cdot cos^2\left(\frac{\pi}{5}\right)-2\cdot cos\left(\frac{\pi}{5}\right)-1\right\rbrack\)

Dựng ΔABC cân tại A, \(\hat{BAC}=36^0\) ; BC=1

Gọi BD là phân giác của góc ABC(D∈AC)

ΔABC cân tại A

=>\(\hat{ABC}=\hat{ACB}=\frac{180^0-\hat{BAC}}{2}=\frac{180^0-36^0}{2}=72^0\)

BD là phân giác của góc ABC

=>\(\hat{ABD}=\hat{DBC}=\frac12\cdot\hat{ABC}=36^0\)

Xét ΔBDC có \(\hat{BDC}+\hat{BCD}+\hat{DBC}=180^0\)

=>\(\hat{BDC}=180^0-36^0-72^0=72^0\)

Xét ΔDAB có \(\hat{DAB}=\hat{DBA}\left(=36^0\right)\)

nên ΔDAB cân tại D

=>DA=DB

Xét ΔBDC có \(\hat{BDC}=\hat{BCD}=72^0\)

nên ΔBDC cân tại B

=>BD=BC=1

=>DA=DB=BC=1

Kẻ DH⊥AB tại H

ΔDAB cân tại D

mà DH là đường cao

nên H là trung điểm của AB

=>HA=HB=x

Xét ΔHAD vuông tại H có cos A\(=\frac{AH}{AD}=x\)

=>\(cosA=\frac{x}{AD}=x\)

DA+DC=AC

=>DC=AC-DA=AB-DA=2x-1

AC=AD+DC=1+2x-1=2x

=>AB=2x

Xét ΔBAC có BD là phân giác

nên \(\frac{DC}{DA}=\frac{BC}{BA}\)

=>\(\frac{2x-1}{1}=\frac{1}{2x}\)

=>2x(2x-1)=1

=>\(4x^2-2x-1=0\)

=>\(x^2-\frac12x-\frac14=0\)

=>\(x^2-2\cdot x\cdot\frac14+\frac{1}{16}-\frac{5}{16}=0\)

=>\(\left(x-\frac14\right)^2=\frac{5}{16}\)

=>\(x-\frac14=\frac{\sqrt5}{4}\)

=>\(x=\frac{\sqrt5+1}{4}\)

=>\(cos36=\frac{\sqrt5+1}{4}\)

=>\(cos\left(\frac{\pi}{5}\right)=\frac{\sqrt5+1}{4}\)

\(4\cdot cos^2\left(\frac{\pi}{5}\right)-2\cdot cos\left(\frac{\pi}{5}\right)-1\)

\(\)\(=4\cdot\left(\frac{\sqrt5+1}{4}\right)^2-2\cdot\frac{\sqrt5+1}{4}-1\)

\(=\frac{4\cdot\left(6+2\sqrt5\right)}{16}-\frac{\sqrt5+1}{2}-1=\frac{8\left(3+\sqrt5\right)}{16}-\frac{\sqrt5+1}{2}-1\)

\(=\frac{3+\sqrt5}{2}-\frac{\sqrt5+1}{2}-1=\frac{3+\sqrt5-\sqrt5-1}{2}-1=\frac22-1=0\)

=>Q=0

=>Q không phụ thuộc vào biến x

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=-\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{4}=\pi-\left(-\dfrac{\pi}{6}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{6}-\dfrac{\pi}{4}+k2\pi\\2x=\pi+\dfrac{\pi}{6}-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{5\pi}{12}+k2\pi\\2x=\dfrac{11\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5\pi}{24}+k\pi\\x=\dfrac{11\pi}{24}+k\pi\end{matrix}\right.\left(k\in Z\right)\)

=)) sẽ chỉ có 1 người trl:vv