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1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
a.
\(1+sin\left(3x-\frac{\pi}{3}\right)=sin^2x+cos^2x+2sinx.cosx\)
\(\Leftrightarrow1+sin\left(3x-\frac{\pi}{3}\right)=1+sin2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=2x+k2\pi\\3x-\frac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(sin2x+cos2x=\frac{\sqrt{6}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{6}}{2}\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\2x+\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))
\(\left(sin\dfrac{x}{2}-cox\dfrac{x}{2}\right)^2+\sqrt{3}cosx=2sin5x+1\)
⇔\(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}-2sin\dfrac{x}{2}cos\dfrac{x}{2}+\sqrt{3}cosx=2sin5x+1\)
⇔\(1-sinx+\sqrt{3}cosx=2sin5x+1\)
⇔\(sin\left(\dfrac{\Pi}{3}-x\right)=sin5x\)
\(2sinx\left(\sqrt{3}cosx+sinx+2sin3x\right)=1\)
⇔\(2\sqrt{3}sinxcosx+2sin^2x+4sinxsin3x=1\)
⇔\(\sqrt{3}sin2x+1-cos2x+cos2x-2cos4x=1\)
⇔\(\sqrt{3}sin2x+cos2x=2cos4x\)
⇔\(cos\left(2x-\dfrac{\Pi}{3}\right)=cos4x\)
Câu a tiếp tục ko dịch được đề :)
b.
\(\Leftrightarrow1+cos3x=sin^2\frac{x}{2}+cos^2\frac{x}{2}+2sin\frac{x}{2}.cos\frac{x}{2}\)
\(\Leftrightarrow1+cos3x=1+sinx\)
\(\Leftrightarrow cos3x=sinx\)
\(\Leftrightarrow cos3x=cos\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow...\)
c.
\(\Leftrightarrow cos8x+cos2x+sinx=cos8x\)
\(\Leftrightarrow cos2x+sinx=0\)
\(\Leftrightarrow cos2x=-sinx\)
\(\Leftrightarrow cos2x=cos\left(\frac{\pi}{2}+x\right)\)
\(\Leftrightarrow...\)
d.
\(sin\left(3x-\frac{\pi}{6}\right)=-\left(1-2sin^2\frac{x}{2}\right)\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=-cosx\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=sin\left(x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow...\)
Mình biết làm r ạ, cảm ơn