Bài 1:
1: \(cosx-\sqrt3\cdot\sin x=2\)
=>\(\frac12\cdot cosx-\frac{\sqrt3}{2}\cdot\sin x=1\)
=>\(\sin\left(\frac{\pi}{6}-x\right)=1\)
=>\(\frac{\pi}{6}-x=\frac{\pi}{2}+k2\pi\)
=>\(x=\frac{\pi}{6}-\frac{\pi}{2}-k2\pi=-\frac23\pi-k2\pi\)
2: \(\sqrt2\cdot cosx+\sqrt2\cdot\sin x=1\)
=>\(\sqrt2\cdot\left(\sin x+cosx\right)=1\)
=>\(\sqrt2\cdot\sqrt2\cdot\sin\left(x+\frac{\pi}{4}\right)=1\)
=>\(\sin\left(x+\frac{\pi}{4}\right)=\frac12\)
=>\(\left[\begin{array}{l}x+\frac{\pi}{4}=\frac{\pi}{6}+k2\pi\\ x+\frac{\pi}{4}=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}-\frac{\pi}{4}+k2\pi=-\frac{\pi}{12}+k2\pi\\ x=\frac56\pi-\frac{\pi}{4}+k2\pi=\frac{7}{12}\pi+k2\pi\end{array}\right.\)
3: \(\sqrt3\cdot cos4x-\sin4x=-1\)
=>\(\frac{\sqrt3}{2}\cdot cos4x-\frac12\cdot\sin4x=-\frac12\)
=>\(\sin\left(\frac{\pi}{3}-4x\right)=-\frac12\)
=>\(\left[\begin{array}{l}\frac{\pi}{3}-4x=-\frac{\pi}{6}+k2\pi\\ \frac{\pi}{3}-4x=\pi+\frac{\pi}{6}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=\frac{\pi}{3}+\frac{\pi}{6}-k2\pi=\frac{\pi}{2}-k2\pi\\ 4x=\frac{\pi}{3}-\frac76\pi-k2\pi=-\frac56\pi-k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{8}-\frac{k\pi}{2}\\ x=-\frac{5}{24}\pi-\frac{k\pi}{2}\end{array}\right.\)
4: \(cos\left(x+\frac{7\pi}{2}\right)-\sqrt3\cdot\sin x=1\)
=>\(cos\left(x+4\pi-\frac{\pi}{2}\right)-\sqrt3\cdot\sin x=1\)
=>\(cos\left(x-\frac{\pi}{2}\right)-\sqrt3\cdot\sin x=1\)
=>\(\sin x-\sqrt3\cdot\sin x=1\)
=>\(\sin x\left(1-\sqrt3\right)=1\)
=>\(\sin x=\frac{1}{1-\sqrt3}\)
=>x∈∅
Bài 2:
\(cos3x-\sqrt3\cdot\sin x=\sqrt3\cdot\sin3x-cosx\)
=>\(cos3x-\sqrt3\cdot\sin3x=\sqrt3\cdot\sin x-cosx\)
=>\(\frac12\cdot cos3x-\frac{\sqrt3}{2}\cdot\sin3x=\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx\)
=>\(\sin\left(\frac{\pi}{6}-3x\right)=\sin\left(x-\frac{\pi}{6}\right)\)
=>\(\left[\begin{array}{l}\frac{\pi}{6}-3x=x-\frac{\pi}{6}+k2\pi\\ \frac{\pi}{6}-3x=\pi-x+\frac{\pi}{6}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}-4x=-\frac{\pi}{3}+k2\pi\\ -2x=\pi+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{\pi}{12}-\frac{k\pi}{2}\\ x=-\frac{\pi}{2}-k\pi\end{array}\right.\)
TH1: \(x=\frac{\pi}{12}-\frac{k\pi}{2}\)
\(x\in\left\lbrack-\pi;\pi\right\rbrack\)
=>\(\frac{\pi}{12}-\frac{k\pi}{2}\in\left\lbrack-\pi;\pi\right\rbrack\)
=>\(\frac{1}{12}-\frac{k}{2}\in\left\lbrack-1;1\right\rbrack\)
=>\(-\frac{k}{2}\in\left\lbrack-\frac{13}{12};\frac{11}{12}\right\rbrack\)
=>k∈[13/6;-11/6]
mà k nguyên
nên k∈{2;1;0;-1}
Khi k=2 thì \(x=\frac{\pi}{12}-\frac{k\pi}{2}=\frac{\pi}{12}-\frac{2\pi}{2}=\frac{\pi}{12}-\pi=-\frac{11}{12}\pi\)
Khi k=1 thì \(x=\frac{\pi}{12}-\frac{k\pi}{2}=\frac{\pi}{12}-\frac{\pi}{2}=-\frac{5}{12}\pi\)
Khi k=0 thì \(x=\frac{\pi}{12}-\frac{k\pi}{2}=\frac{\pi}{12}\)
Khi k=-1 thì \(x=\frac{\pi}{12}-\frac{k\pi}{2}=\frac{\pi}{12}+\frac{\pi}{2}=\frac{7}{12}\pi\)
TH2: \(x=-\frac{\pi}{2}-k\pi\)
\(x\in\left\lbrack-\pi;\pi\right\rbrack\)
=>\(-\frac{\pi}{2}-k\pi\in\left\lbrack-\pi;\pi\right\rbrack\)
=>\(k+\frac12\in\left\lbrack-1;1\right\rbrack\)
=>k∈[-3/2;1/2]
mà k nguyên
nên k∈{-1;0}
Khi k=-1 thì \(x=-\frac{\pi}{2}-k\pi=-\frac{\pi}{2}-\left(-1\right)\cdot\pi=-\frac{\pi}{2}+\pi=\frac{\pi}{2}\)
Khi k=0 thì \(x=-\frac{\pi}{2}-k\pi=-\frac{\pi}{2}\)
