`cos 2x+\sqrt{3}sin 2x+\sqrt{3}sin x-cos x=4`

`<=>1/2 cos 2x+\sqrt{3}/2 sin 2x+\sqrt{3}/2 sin x-1/2 cos x=2`

`<=>sin(\pi/6 +2x)+sin(x-\pi/6)=2`

Vì `-1 <= sin (\pi/6 +2x) <= 1`

     `-1 <= sin (x-\pi/6) <= 1`

 Dấu "`=`" xảy ra `<=>{(sin(\pi/6+2x)=1),(sin(x-\pi/6)=1):}`

        `<=>{(\pi/6+2x=\pi/2+k2\pi),(x-\pi/6=\pi/2+k2\pi):}`

        `<=>{(x=\pi/6+k\pi),(x=[2\pi]/3+k2\pi):}`    `(k in ZZ)`

Em cảm ơn ạ.

hộ vs ae ơi

sao ra được dòng thứ 2 vậy bạn

a: \(\frac{1}{\sin x}+\frac{1}{cosx}=4\cdot\sin\left(x+\frac{\pi}{4}\right)\)

=>\(\frac{\sin x+cosx}{\sin x\cdot cosx}=4\cdot\frac{\sqrt2}{2}\cdot\left(\sin x+cosx\right)\)

=>\(\left(\sin x+cosx\right)\left(\frac{1}{\sin x\cdot cosx}-2\sqrt2\right)=0\)

TH1: \(\frac{1}{\sin x\cdot cosx}-2\sqrt2=0\)

=>\(\frac{1}{\sin x\cdot cosx}=2\sqrt2\)

=>\(sinx\cdot cosx=\frac{1}{2\sqrt2}\)

=>\(2\cdot\sin x\cdot cosx=\frac{1}{\sqrt2}\)

=>\(\sin2x=\frac{1}{\sqrt2}\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{4}+k2\pi\\ 2x=-\frac{\pi}{4}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{8}+k\pi\\ x=-\frac{\pi}{8}+k\pi\end{array}\right.\)

TH2: sin x+cosx=0

=>\(\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(\sin\left(x+\frac{\pi}{4}\right)=0\)

=>\(x+\frac{\pi}{4}=k\pi\)

=>\(x=-\frac{\pi}{4}+k\pi\)

\(\sqrt{2}sinx+sin2x=\sqrt{3}cos2x-\sqrt{6}cosx\)

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sinx+\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x+\dfrac{\sqrt{6}}{2}cosx=0\)

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

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{6}\right)+2sin\left(x-\dfrac{\pi}{6}\right).cos\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{6}\right)\left[1+\sqrt{2}sin\left(x-\dfrac{\pi}{6}\right)\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=-\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)

Đến đấy thì dễ rồi.

\(\Leftrightarrow\sqrt{2}\left(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x=0\)

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

Đặt \(x+\dfrac{\pi}{3}=u\Rightarrow2x-\dfrac{\pi}{3}=2u-\pi\)

\(\Rightarrow\sqrt{2}sinu+sin\left(2u-\pi\right)=0\)

\(\Leftrightarrow\sqrt{2}sinu-sin2u=0\)

\(\Leftrightarrow sinu\left(\sqrt{2}-2cosu\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=0\\cosu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=0\\cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

Ta có: \(\sin2x-cosx=\sqrt3\left(\sin x+cos2x\right)\)

=>\(\sin2x-\sqrt3\cdot cos2x=\sqrt3\cdot\sin x+cosx\)

=>\(\frac12\cdot\sin2x-\frac{\sqrt3}{2}\cdot cos2x=\frac{\sqrt3}{2}\cdot\sin x+\frac12\cdot cosx\)

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

=>\(\left[\begin{array}{l}2x-\frac{\pi}{3}=x+\frac{\pi}{6}+k2\pi\\ 2x-\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi=-x+\frac56\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}2x-x=\frac{\pi}{6}+\frac{\pi}{3}+k2\pi=\frac{\pi}{2}+k2\pi\\ 2x+x=\frac56\pi+\frac{\pi}{3}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{2}+k2\pi\\ x=\frac{7}{18}\pi+\frac{k2\pi}{3}\end{array}\right.\)

1.

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

\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)

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

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

Xét (1):

Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

2.

\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)

\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)

Xét (1):

Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm

3.

\(4sinx.cosx-2sinx+1-2cosx=0\)

\(\Leftrightarrow2sinx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\)

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

4.

\(cosx-sinx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\-4sinx.cosx=2t^2-2\end{matrix}\right.\)

Pt trở thành: \(t+2t^2-2-1=0\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{3}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)

5.

\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=sinx\)

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

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

\(\Leftrightarrow...\)

6.

\(9sin^2x-5\left(1-sin^2x\right)-5sinx+4=0\)

\(\Leftrightarrow14sin^2x-5sinx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(-\frac{1}{7}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{7}\right)+k2\pi\end{matrix}\right.\)