a: sin 3x-cos3x+\(\sqrt3=0\)

=>\(\sin3x-cos3x=-\sqrt3\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=-\sqrt3\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=-\sqrt{\frac32}<-1\)

=>Phương trình không có nghiệm

b: sin x=căn 2

mà căn 2>1

nên x∈∅

=>Tập nghiệm là S=∅

c: \(\sin2x=\frac{\sqrt3}{2}\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\\ 2x=\pi-\frac{\pi}{3}+k2\pi=\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+k\pi\\ x=\frac{\pi}{3}+k\pi\end{array}\right.\)

TH1: \(x=\frac{\pi}{6}+k\pi\)

\(x\in\left\lbrack-\pi;2\pi\right\rbrack\)

=>\(\frac{\pi}{6}+k\pi\in\left\lbrack-\pi;2\pi\right\rbrack\)

=>\(k+\frac16\in\left\lbrack-1;2\right\rbrack\)

=>\(k\in\left\lbrack-\frac76;\frac{11}{6}\right\rbrack\)

mà k nguyên

nên k∈{-1;0;1}

=>Có 3 nghiệm trong trường hợp này(1)

TH2: \(x=\frac{\pi}{3}+k\pi\)

x\(\in\left\lbrack-\pi;2\pi\right\rbrack\)

=>\(\frac{\pi}{3}+k\pi\in\left\lbrack-\pi;2\pi\right\rbrack\)

=>\(k+\frac13\in\left\lbrack-1;2\right\rbrack\)

=>k∈[-4/3;5/3]

mà k nguyên

nên k∈{-1;0;1}

=>Có 3 nghiệm trong trường hợp này(2)

Từ (1),(2) suy ra có 3+3=6 nghiệm

\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)

\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)

\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)

b) vì \(\sqrt{5}-\sqrt{12}< 0\), ta có: 

 \(5\sqrt{5}-2\sqrt{3}=4\sqrt{5}+\sqrt{5}-\sqrt{12}< 4\sqrt{5}< 4\sqrt{5}+6\) 

Vậy \(5\sqrt{5}-2\sqrt{3}< 6+4\sqrt{5}\)

\(cos^2x-\left(2sin\frac{x}{2}cos\frac{x}{2}\right)^2=cos^2x-sin^2x=cos2x\)

\(\frac{sin3x}{sinx}-\frac{cos3x}{cosx}=\frac{sin3x.cosx-cos3x.sinx}{sinx.cosx}=\frac{sin\left(3x-x\right)}{\frac{1}{2}sin2x}=\frac{2sin2x}{sin2x}=2\)

\(\frac{cosx+cos3x+cos2x+cos4x}{sinx+sin3x+sin2x+sin4x}=\frac{2cosx.cos2x+2cosx.cos3x}{2sin2x.cosx+2sin3x.cosx}=\frac{2cosx\left(cos2x+cos3x\right)}{2cosx\left(sin2x+sin3x\right)}\)

\(=\frac{cos2x+cos3x}{sin2x+sin3x}=\frac{2cos\frac{x}{2}.cos\frac{5x}{2}}{2sin\frac{5x}{2}.cos\frac{x}{2}}=cot\frac{5x}{2}\)