\(3sin3x-4sin^33x+\sqrt{3}sin9x=1\)

\(\Leftrightarrow sin9x+\sqrt{3}sin9x=1\)

\(\Leftrightarrow\left(\sqrt{3}+1\right)sin9x=1\)

\(\Leftrightarrow sin9x=\frac{1}{\sqrt{3}+1}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{9}arcsin\left(\frac{1}{\sqrt{3}+1}\right)+k2\pi\\x=\pi-\frac{1}{9}arcsin\left(\frac{1}{\sqrt{3}+1}\right)+k2\pi\end{matrix}\right.\)

f: \(cos7x-\sqrt3\cdot\sin7x-\sin x=\sqrt3\cdot cosx\)

=>\(\frac12\cdot cos7x-\frac{\sqrt3}{2}\cdot\sin7x=\frac12\cdot\sin x+\frac{\sqrt3}{2}\cdot cosx\)

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

=>\(\left[\begin{array}{l}-7x+\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\ -7x+\frac{\pi}{6}=\pi-x-\frac{\pi}{3}+k2\pi=-x+\frac23\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}-7x-x=\frac{\pi}{3}-\frac{\pi}{6}+k2\pi=\frac{\pi}{6}+k2\pi\\ -7x+x=\frac23\pi-\frac{\pi}{6}+k2\pi=\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}-8x=\frac{\pi}{6}+k2\pi\\ -6x=\frac12\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{\pi}{48}-\frac{k\pi}{4}\\ x=-\frac{1}{12}\pi-\frac{k\pi}{3}\end{array}\right.\)

e: \(5\cdot\sin2x-6\cdot cos^2x=13\)

=>\(5\cdot\sin2x-6\cdot\frac{1+cos2x}{2}=13\)

=>\(5\cdot\sin2x-3-3\cdot cos2x=13\)

=>\(5\cdot\sin2x-3\cdot cos2x=16\)

Vì \(5^2+\left(-3\right)^2=25+9=34<16^2\)

nên phương trình vô nghiệm

Câu 1:

\(cos^2\) gì nhỉ?

Câu 2: đề không hợp lý \(\sqrt{3}sin9x\) là \(\sqrt{3}cos9x\) có lý hơn

\(\Leftrightarrow3sin3x-4sin^33x+\sqrt{3}sin9x=1\)

\(\Leftrightarrow sin9x+\sqrt{3}sin9x=1\)

\(\Leftrightarrow sin9x=\frac{1}{\sqrt{3}+1}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{9}arcsin\left(\frac{1}{\sqrt{3}+1}\right)+\frac{k2\pi}{9}\\x=\frac{\pi}{9}-\frac{1}{9}arcsin\left(\frac{1}{\sqrt{3}+1}\right)+\frac{k2\pi}{9}\end{matrix}\right.\)

Nghiệm nhìn rất ngớ ngẩn nếu đề đúng

3.

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=\frac{1}{2}\)

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

\(\Leftrightarrow...\)

Chọn D

Vậy phương trình có 5 nghiệm thỏa mãn.

Có b nào gipus mk với cần gấp gấp :)

\(4sin^2x.cosx+2cos2x=cosx+\sqrt{3}sin3x\)

\(\Leftrightarrow2\left(1-cos2x\right).cosx+2cos2x=cosx+\sqrt{3}sin3x\)

\(\Leftrightarrow2cosx-2cos2x.cosx+2cos2x=cosx+\sqrt{3}sin3x\)

\(\Leftrightarrow2cosx-cos3x-cosx+2cos2x=cosx+\sqrt{3}sin3x\)

\(\Leftrightarrow\sqrt{3}sin3x+cos3x=2cos2x\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin3x+\dfrac{1}{2}cos3x=cos2x\)

\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos2x\)

\(\Leftrightarrow3x-\dfrac{\pi}{3}=\pm2x+k2\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{15}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)