\(sin3x+\sqrt{3}cos3x=2sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}sin3x+\frac{\sqrt{3}}{2}cos3x\)

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

\(\Leftrightarrow x\in R\)

Đề là \(sin\left(5-\dfrac{\pi}{3}\right)\) hay \(sin\left(5x-\dfrac{\pi}{3}\right)\) nhỉ?

\(sin3x-cos3x=\sqrt{2}sin\left(5x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=\sqrt{2}sin\left(5x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(5x-\dfrac{\pi}{3}\right)=sin\left(3x-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{\pi}{3}=3x-\dfrac{\pi}{4}+k2\pi\\5x-\dfrac{\pi}{3}=\dfrac{5\pi}{4}-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{24}+k\pi\\x=\dfrac{19\pi}{96}+\dfrac{k\pi}{4}\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}.sin\left(3x-\dfrac{\pi}{4}\right)-\sqrt{2}.sin\left(5x-\dfrac{\pi}{3}\right)=0\Leftrightarrow sin\left(3x-\dfrac{\pi}{4}\right)=sin\left(5x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{4}+k2\pi=5x-\dfrac{\pi}{3}\\\pi-3x+\dfrac{\pi}{4}+k2\pi=5x-\dfrac{\pi}{3}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{19\pi}{96}+\dfrac{k\pi}{4}\end{matrix}\right.\); k\(\in Z\)

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 \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

ĐKXĐ:...

Biến đổi đoạn trong ngoặc trước cho đỡ rối:

\(cos4x+sin2x=cos\left(3x+x\right)+sin\left(3x-x\right)\)

\(=cos3x.cosx-sin3x.sinx+sin3x.cosx-cos3x.sinx\)

\(=cosx\left(cos3x+sin3x\right)-sinx\left(cos3x+sin3x\right)\)

\(=\left(cosx-sinx\right)\left(cos3x+sin3x\right)\)

Thay vào phương trình:

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

\(\Leftrightarrow1-2sinx.cosx=2\left(sinx+cosx\right)+3\)

Đặt \(sinx+cosx=a\Rightarrow-2sinx.cosx=1-a^2\)

\(2-a^2=2a+3\Rightarrow a=-1\Rightarrow sinx+cosx=-1\Rightarrow...\)

Hình như câu này tui từng đi hỏi anh Lâm thì phải :D

\(\sin3x+\cos3x=3\sin x-4\sin^3x+4\cos^3x-3\cos x\)

\(=3\left(\sin x-\cos x\right)-4\left(\sin x-\cos x\right)\left(\sin^2x+\sin x\cos x+\cos^2x\right)=\left(\cos x-\sin x\right)\left(4\sin x\cos x+1\right)=\left(\cos x-\sin x\right)\left(1+2\sin2x\right)\)

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

Bạn tự giải nốt, nhớ đối chiếu đkxd nhó