\(\Leftrightarrow\sin\left(2x-10^o\right)=\sin\left(-60^o\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-10^o=-60^o+k360^o\\2x-10^o=180^o-\left(-60^o\right)+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-50^o+k360^o\\2x=250^o+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-25^o+k180^o\\x=125^o+k180^o\end{matrix}\right.\left(k\in Z\right)\)

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

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

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

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

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

\(\Leftrightarrow...\)

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

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

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

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

`2sin(2x+\pi/4)=\sqrt{2}`

`<=>sin(2x+\pi/4)=\sqrt{2}/2`

`<=>` $\left[\begin{matrix} 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\ 2x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}\end{matrix}\right.$    `(k in ZZ)`

`<=>` $\left[\begin{matrix} 2x=k2\pi\\ 2x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.$  `(k in ZZ)`

`<=>` $\left[\begin{matrix} x=k\pi\\ x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.$   `(k in ZZ)`

ĐKXĐ: \(sinx\ne\dfrac{\sqrt{2}}{2}\)

\(\left(sinx-cosx\right)\left(sin2x-3\right)+\left(sinx-cosx\right)^2+\left(sin^2x-cos^2x\right)=0\)

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

\(\Leftrightarrow\left(sinx-cosx\right)\left(sin2x-3+2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\\left(sin2x-1\right)+2\left(sinx+1\right)=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)

Kết hợp ĐKXĐ \(\Rightarrow x=-\dfrac{\pi}{4}+k2\pi\)

Lời giải:ĐK: $\cos 3x>\frac{-1}{2}$

PT $\Rightarrow 4\sin ^2\frac{x}{2}-\sqrt{3}\cos 2x-1-2\cos ^2(x-\frac{3\pi}{4})=0$

$\Leftrightarrow 2(1-\cos x)-\sqrt{3}\cos 2x-2+[1-2\cos ^2(x-\frac{3\pi}{4})]=0$

$\Leftrightarrow -2\cos x-\sqrt{3}\cos 2x-cos (2x-\frac{3\pi}{2})=0$

$\Leftrightarrow 2\cos x+\sqrt{3}\cos 2x+\cos (2x-\frac{3\pi}{2})=0$

$\Leftrightarrow 2\cos x+\sqrt{3}\cos 2x+\sin 2x=0$

$\Leftrightarrow \cos x+\frac{\sqrt{3}}{2}\cos 2x+\frac{1}{2}\sin 2x=0$

$\Leftrightarrow \cos x-\cos (2x+\frac{5\pi}{6})=0

$\Leftrightarrow \cos x=\cos (2x+\frac{5\pi}{6})$

$\Rightarrow x+2k\pi =2x+\frac{5}{6}\pi$ hoặc $-x+2k\pi =2x+\frac{5}{6}\pi$

Vậy......

a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)

a, Ta có : \(\sin\left(3x+60\right)=\dfrac{1}{2}\)

\(\Rightarrow3x+60=30+2k180\)

\(\Rightarrow3x=2k180-30\)

\(\Leftrightarrow x=120k-10\)

Vậy ...

b, Ta có : \(\cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow2x-\dfrac{\pi}{3}=\dfrac{3}{4}\pi+k2\pi\)

\(\Leftrightarrow x=\dfrac{13}{24}\pi+k\pi\)

Vậy ...

c, Ta có : \(tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)

\(\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

Vậy ...

d, Ta có : \(\cot\left(2x+\pi\right)=-1\)

\(\Rightarrow2x+\pi=\dfrac{3}{4}\pi+k\pi\)

\(\Leftrightarrow x=-\dfrac{1}{8}\pi+\dfrac{k}{2}\pi\)

Vậy ...

a) \(sin\left(3x+60^0\right)=\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))

Vậy...

b) Pt\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\dfrac{3\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

c) Pt \(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=tan\dfrac{\pi}{3}\)

\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi,k\in Z\)\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi,k\in Z\)

Vậy...

d) Pt \(\Leftrightarrow tan\left(2x+\pi\right)=-1\)

\(\Leftrightarrow2x+\pi=-\dfrac{\pi}{4}+k\pi,k\in Z\)

\(\Leftrightarrow x=-\dfrac{5\pi}{8}+\dfrac{k\pi}{2},k\in Z\)

Vậy...

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

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

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

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

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

=)) sẽ chỉ có 1 người trl:vv

a.

\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)

\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)

\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)

\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)

\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)

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

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

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)

\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)

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

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

\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)

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

Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)

c.

\(2\sqrt{2}cos\left(\dfrac{5\pi}{12}-x\right)sinx=1\)

\(\Leftrightarrow\sqrt{2}\left(sin\left(\dfrac{5\pi}{12}\right)+sin\left(2x-\dfrac{5\pi}{12}\right)\right)=1\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=\dfrac{-\sqrt{6}+\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{5\pi}{12}\right)=sin\left(-\dfrac{\pi}{12}\right)\)

\(\Leftrightarrow...\)