\(y'=4cos\left(4x-\dfrac{\pi}{3}\right)cos5x-5sin5x.sin\left(4x-\dfrac{\pi}{3}\right)\)

\(\sqrt{3}cos5x-2sin3x.cos2x-sinx=0\)

⇔ \(\sqrt{3}cos5x-\left(sinx+sin5x\right)-sinx=0\)

⇔ \(\sqrt{3}cos5x-sin5x-2sinx=0\)

⇔ \(2sin\left(\dfrac{\pi}{3}-5x\right)=2sinx\)

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

Giải nốt nhé

\(\sqrt{3}\)Cos5x - 2Sin3x.Cos2x - Sinx =0

\(\Leftrightarrow\sqrt{3}\)Cos5x -(Sin5x +Sinx ) -Sinx =0

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}\)Cos5x - \(\dfrac{1}{2}\)Sin5x =Sinx

\(\Leftrightarrow\)Sin (\(\dfrac{\pi}{3}\) - 5x )=Sinx

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\pi}{18}-k\dfrac{\pi}{3}\\x=\dfrac{-\pi}{6}-k\dfrac{\pi}{2}\end{matrix}\right.\)

\(\Rightarrow\) Vậy pt có 2 ng x=\(\dfrac{\pi}{18}-k\dfrac{\pi}{3}\)và x=\(\dfrac{-\pi}{6}-k\dfrac{\pi}{2}\),k\(\varepsilon\)Z

107.

\(\Leftrightarrow tan2x=-\sqrt{3}\)

\(\Leftrightarrow2x=-\frac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=-\frac{\pi}{6}+\frac{k\pi}{2}\)

\(2000\pi\le-\frac{\pi}{6}+\frac{k\pi}{2}\le2018\pi\)

\(\Leftrightarrow4000+\frac{1}{3}\le k\le4036+\frac{1}{2}\)

Có \(4036-4001+1=36\) nghiệm

108.

\(\Leftrightarrow\left[{}\begin{matrix}5x=\frac{\pi}{4}+k2\pi\\5x=-\frac{\pi}{4}+n2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{20}+\frac{k2\pi}{5}\\x=-\frac{\pi}{20}+\frac{n2\pi}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-50\pi\le\frac{\pi}{20}+\frac{k2\pi}{5}\le0\\-50\pi\le-\frac{\pi}{20}+\frac{n2\pi}{5}\le0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-125-\frac{1}{8}\le k\le-\frac{1}{8}\\-125+\frac{1}{8}\le n\le\frac{1}{8}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-125\le k\le-1\\-124\le n\le0\end{matrix}\right.\)

Có \(-1-\left(-125\right)+1+0-\left(-124\right)+1=250\) nghiệm

109.

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0< -\frac{\pi}{12}+k\pi< \pi\\0< \frac{7\pi}{12}+k\pi< \pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{12}< k< \frac{13}{12}\\-\frac{7}{12}< k< \frac{5}{12}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}k=1\\k=0\end{matrix}\right.\) có 2 nghiệm

110.

\(\Leftrightarrow cos2x=-\frac{1}{2}\)

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

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

Ko có đáp án chọn nên ko thể bấm được, chỉ giải được tự luận thôi :)

\(=\dfrac{1}{3}\lim\limits_{x\rightarrow0}\dfrac{sinx}{x}-\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3}cos5x}{3x}=\dfrac{1}{3}-\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3}cos5x}{3x}\)

Xét:

\(\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt{3}cos5x}{3x}=\dfrac{\sqrt{3}}{0}=+\infty\)

\(\lim\limits_{x\rightarrow0^-}\dfrac{-\sqrt{3}cos5x}{-3x}=\dfrac{-\sqrt{3}}{0}=-\infty\)

\(\Rightarrow\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3}cos5x}{3x}\) ko tồn tại nên giới hạn đã cho không tồn tại

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.

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

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

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

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

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