1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

a/

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin5x-\frac{1}{2}cos5x=-1\)

\(\Leftrightarrow sin\left(5x-\frac{\pi}{6}\right)=-1\)

\(\Leftrightarrow5x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\frac{\pi}{15}+\frac{k2\pi}{5}\)

b/

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

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{\sqrt{2}}\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)

d/

\(\Leftrightarrow2sinx.cosx+1-2sin^2x=1\)

\(\Leftrightarrow2sinx\left(cosx-sinx\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

a, \(sin^2x-4sinx+3=0\)

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

\(\Leftrightarrow sinx=1\)

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

b, \(2cos^2-cosx-1=0\)

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

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

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

c, \(3sin^2x-2cosx+2=0\)

\(\Leftrightarrow3-3sin^2x+2cosx-5=0\)

\(\Leftrightarrow3cos^2x+2cosx-5=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(3cosx+5\right)=0\)

\(\Leftrightarrow cosx=1\)

\(\Leftrightarrow x=k2\pi\)

1.

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

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{3}=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+k\pi\\x=-\frac{7\pi}{24}+k\pi\end{matrix}\right.\)

2.

\(2\left(1-cosx\right)-3\sqrt{3}sinx-\left(1+cosx\right)=4\)

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

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

\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=-\frac{1}{2}\)

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

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.

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

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

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

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

hộ vs ae ơi

1.

\(\Leftrightarrow sin5x+\sqrt{3}cos5x=-2sin15x\)

\(\Leftrightarrow\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x=-sin15x\)

\(\Leftrightarrow sin\left(5x+\frac{\pi}{3}\right)=sin\left(-15x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=-15x+k2\pi\\5x+\frac{\pi}{3}=\pi+15x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{60}+\frac{k\pi}{10}\\x=-\frac{\pi}{15}+\frac{k\pi}{5}\end{matrix}\right.\)

2.

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

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

Do \(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)\le1\\sin\left(x+\frac{\pi}{6}\right)\le1\end{matrix}\right.\) với mọi x

\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)+sin\left(x+\frac{\pi}{6}\right)\le2\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)=1\\sin\left(x+\frac{\pi}{6}\right)=1\end{matrix}\right.\)

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

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

\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)

3.

\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)

a/

\(\Leftrightarrow4cos^2\left(6x-2\right)+8\left(1+cos\left(6x-2\right)\right)-13=0\)

Đặt \(cos\left(6x-2\right)=a\Rightarrow\left|a\right|\le1\)

Pt trở thành:

\(4a^2+8\left(1+a\right)-13=0\)

\(\Leftrightarrow4a^2+8a-5=0\Rightarrow\left[{}\begin{matrix}a=\frac{1}{2}\\a=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cos\left(6x-2\right)=\frac{1}{2}\)

\(\Rightarrow6x-2=\pm\frac{\pi}{3}+k2\pi\)

\(\Rightarrow x=\frac{1}{3}\pm\frac{\pi}{18}+\frac{k\pi}{3}\)

b/

\(\Leftrightarrow2cos^2\left(x+75^0\right)-1+3sin\left(15^0-x\right)-1=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(90^0-15^0+x\right)-2=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(x+75^0\right)-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cos\left(x+75^0\right)=\frac{1}{2}\\cos\left(x+75^0\right)=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+75^0=60^0+k360^0\\x+75^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+k360^0\\x=-135^0+k360^0\end{matrix}\right.\)

c/

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

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow cos2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow1-2sin^2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)\left(1-2sin\left(x-\frac{\pi}{6}\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{6}\right)=0\\sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=k\pi\\x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)