\(\Leftrightarrow\frac{\sqrt{3}}{2}.\left(3\sin x-4\sin^3x\right)-\sin x.\cos x=0\)

\(\Leftrightarrow3\sqrt{3}\sin x-4\sqrt{3}\sin^3x-2\sin x.\cos x=0\)

Xét \(\sin x=0\) là nghiệm của pt=> \(x=\pi\)

Xét \(\sin x\ne0\) chia cả 2 vế cho \(\sin x\)

\(\Leftrightarrow3\sqrt{3}-4\sqrt{3}\sin^2x-2\cos x=0\)

\(\Leftrightarrow3\sqrt{3}-4\sqrt{3}+4\sqrt{3}\cos^2x-2\cos x=0\)

\(\Leftrightarrow4\sqrt{3}\cos^2x-2\cos x-\sqrt{3}=0\)

đây là pt b2, bn tự giải và kết luận :3

a.

\(\Leftrightarrow\left[{}\begin{matrix}3x=90^0-x+k360^0\\3x=90^0+x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{45^0}{2}+k90^0\\x=45^0+k180^0\end{matrix}\right.\)

b.

\(\Leftrightarrow cos\left(3x+45^0\right)=cos\left(x-180^0\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+45^0=x-180^0+k360^0\\3x+45^0=180^0-x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{225^0}{2}+k180^0\\x=\frac{135^0}{4}+k90^0\end{matrix}\right.\)

c.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

d.

\(\Leftrightarrow sin\left(x-\frac{2\pi}{3}\right)=cos2x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{12}+k\pi\\x=-\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

e.

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

\(\Leftrightarrow cos\left(2x-\frac{\pi}{4}\right)=cos\left(\frac{\pi}{6}-2x\right)\)

\(\Leftrightarrow2x-\frac{\pi}{4}=\frac{\pi}{6}-2x+k2\pi\)

\(\Leftrightarrow x=\frac{5\pi}{48}+\frac{k\pi}{2}\)

\(\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x+sin\frac{9x}{4}=2\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)+sin\left(\frac{9x}{4}\right)=2\)

Do \(\left\{{}\begin{matrix}sin\left(3x-\frac{\pi}{6}\right)\le1\\sin\left(\frac{9x}{4}\right)\le1\end{matrix}\right.\)

Nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(3x-\frac{\pi}{6}\right)=1\\sin\left(\frac{9x}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2\pi}{9}+\frac{k8\pi}{3}\)

đề câu 1 đúng r

ngại viết quá hihi, mà hơi ngáo tí cái dạng này lm rồi mà cứ quên

bài trước mk bình luận bạn đọc chưa nhỉ

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

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

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

6.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

1: \(cos^2\left(x-\frac{\pi}{5}\right)=\sin^2\left(2x+\frac45\pi\right)\)

=>\(\left[\begin{array}{l}cos\left(x-\frac{\pi}{5}\right)=\sin\left(2x+\frac45\pi\right)=cos\left(\frac{\pi}{2}-2x-\frac45\pi\right)=cos\left(-2x-\frac{3}{10}\pi\right)\\ cos\left(x-\frac{\pi}{5}\right)=-\sin\left(2x+\frac45\pi\right)=\sin\left(-2x-\frac45\pi\right)=cos\left(\frac{\pi}{2}+2x+\frac45\pi\right)=cos\left(2x+\frac{13}{10}\pi\right)\end{array}\right.\)

TH1: \(cos\left(x-\frac{\pi}{5}\right)=cos\left(-2x-\frac{3}{10}\pi\right)\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{5}=-2x-\frac{3}{10}\pi+k2\pi\\ x-\frac{\pi}{5}=2x+\frac{3}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x=-\frac{3}{10}\pi+\frac{\pi}{5}+k2\pi=-\frac{1}{10}\pi+k2\pi\\ -x=\frac{3}{10}\pi+\frac{\pi}{5}+k2\pi=-\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{30}\pi+\frac{k2\pi}{3}\\ x=\frac12\pi-k2\pi\end{array}\right.\)

TH2: \(cos\left(x-\frac{\pi}{5}\right)=cos\left(2x+\frac{13}{10}\pi\right)\)

=>\(\left[\begin{array}{l}2x+\frac{13}{10}\pi=x-\frac{\pi}{5}+k2\pi\\ 2x+\frac{13}{10}\pi=-x+\frac{\pi}{5}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-x=-\frac{\pi}{5}-\frac{13}{10}\pi+k2\pi\\ 2x+x=\frac{\pi}{5}-\frac{13}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{15}{10}\pi+k2\pi=-\frac32\pi+k2\pi\\ 3x=-\frac{11}{10}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac32\pi+k2\pi\\ x=-\frac{11}{30}\pi+\frac{k2\pi}{3}\end{array}\right.\)

2: \(\sin3x=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)+cos3x\)

=>\(\sin3x-cos3x=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=cos\left(x-\frac{\pi}{5}\right)=\sin\left(\frac{\pi}{2}-x+\frac{\pi}{5}\right)=\sin\left(-x+\frac{7}{10}\pi\right)\)

=>\(\left[\begin{array}{l}3x-\frac{\pi}{4}=-x+\frac{7}{10}\pi+k2\pi\\ 3x-\frac{\pi}{4}=\pi+x-\frac{7}{10}\pi+k2\pi=x+\frac{3}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}4x=\frac{7}{10}\pi+\frac{\pi}{4}+k2\pi=\frac{19}{20}\pi+k2\pi\\ 2x=\frac{3}{10}\pi+\frac{\pi}{4}+k2\pi=\frac{11}{20}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{19}{80}\pi+\frac{k\pi}{2}\\ x=\frac{11}{40}\pi+k\pi\end{array}\right.\)

c/

\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)

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

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

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

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

a/

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

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

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

b/

\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)

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

\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)