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

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

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

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

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

\(\Leftrightarrow...\)

a) \(sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=-sin\left(x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{3}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-x+k\pi\\2x+\dfrac{\pi}{6}=\pi-\dfrac{\pi}{3}+x+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

b) \(sin\left(2x-\dfrac{\pi}{3}\right)-cos\left(x+\dfrac{\pi}{3}\right)=0\)

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

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(\dfrac{\pi}{6}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+x+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{7\pi}{6}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{6}+\left(k+1\right)\pi\end{matrix}\right.\)

c: =>\(cos\left(x-\dfrac{pi}{6}\right)=-sin\left(2x+\dfrac{pi}{3}\right)\)

=>\(cos\left(x-\dfrac{pi}{6}\right)=sin\left(-2x-\dfrac{pi}{3}\right)\)

=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(\dfrac{pi}{2}-x+\dfrac{pi}{6}\right)\)

=>\(sin\left(-2x-\dfrac{pi}{3}\right)=sin\left(-x+\dfrac{2}{3}pi\right)\)

=>\(\left[{}\begin{matrix}-2x-\dfrac{pi}{3}=-x+\dfrac{2}{3}pi+k2pi\\-2x-\dfrac{pi}{3}=pi+x-\dfrac{2}{3}pi+k2pi\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}-x=pi+k2pi\\-3x=\dfrac{2}{3}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-pi-k2pi\\x=-\dfrac{2}{9}pi-\dfrac{k2pi}{3}\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

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

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

\(=0\)

1: cos(2x+pi/6)=cos(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=3x-pi/3+k2pi

=>5x=pi/6+k2pi hoặc -x=-1/2pi+k2pi

=>x=pi/30+k2pi/5 hoặc x=pi-k2pi

2: sin(2x+pi/6)=sin(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=pi-pi/3+3x+k2pi

=>5x=pi/6+k2pi hoặc -x=2/3pi-pi/6+k2pi

=>x=pi/30+k2pi/5 hoặc x=-1/2pi-k2pi

1) \(cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{3}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-3x+k2\pi\\2x+\dfrac{\pi}{6}=-\dfrac{\pi}{3}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{3}-\dfrac{\pi}{6}+k2\pi\\3x-2x=\dfrac{\pi}{3}+\dfrac{\pi}{6}-k2\pi\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\) \(\left(k\in N\right)\)

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

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

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

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