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7.
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(\frac{\pi}{4}-x\right).sin\left(\frac{\pi}{4}+x\right)\ne0\\cos\left(\frac{\pi}{4}-x\right)cos\left(\frac{\pi}{4}+x\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow cos2x\ne0\)
Phương trình tương đương:
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{2}-\frac{\pi}{4}-x\right)}=cos^44x\)
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{4}-x\right)}=cos^24x\)
\(\Leftrightarrow sin^42x+cos^42x=cos^44x\)
\(\Leftrightarrow\left(sin^22x+cos^22x\right)^2-2sin^22x.cos^22x=cos^44x\)
\(\Leftrightarrow1-\frac{1}{2}sin^24x=cos^44x\)
\(\Leftrightarrow2-\left(1-cos^24x\right)=2cos^44x\)
\(\Leftrightarrow2cos^44x-cos^24x-1=0\)
\(\Leftrightarrow\left(cos^24x-1\right)\left(2cos^24x+1\right)=0\)
\(\Leftrightarrow cos^24x-1=0\)
\(\Leftrightarrow sin^24x=0\Leftrightarrow sin4x=0\)
\(\Leftrightarrow2sin2x.cos2x=0\Leftrightarrow sin2x=0\)
\(\Leftrightarrow x=\frac{k\pi}{2}\)
1.
\(cos2x+5=2\left(2-cosx\right)\left(sinx-cosx\right)\)
\(\Leftrightarrow2cos^2x+4=4sinx-4cosx-2sinx.cosx+2cos^2x\)
\(\Leftrightarrow2sinx.cosx-4\left(sinx-cosx\right)+4=0\)
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(1-t^2-4t+4=0\)
\(\Leftrightarrow t^2+4t-5=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
Hình như câu này tui từng đi hỏi anh Lâm thì phải :D
\(\sin3x+\cos3x=3\sin x-4\sin^3x+4\cos^3x-3\cos x\)
\(=3\left(\sin x-\cos x\right)-4\left(\sin x-\cos x\right)\left(\sin^2x+\sin x\cos x+\cos^2x\right)=\left(\cos x-\sin x\right)\left(4\sin x\cos x+1\right)=\left(\cos x-\sin x\right)\left(1+2\sin2x\right)\)
\(\Leftrightarrow\sqrt{3}\cos x=\sin x\Leftrightarrow\sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
Bạn tự giải nốt, nhớ đối chiếu đkxd nhó
\(sin3x+\sqrt{3}cos3x=2sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}sin3x+\frac{\sqrt{3}}{2}cos3x\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow x\in R\)
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⇔ .png)
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+ k2π ⇔ x = .png)
, (k ∈ Z)..png)
) = 0 ⇔ .png)
, (k ∈ Z)..png)
= sin(-600) nên phương trình đã cho tương đương với.png)
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a.
\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)
\(tan3x-tanx=0\)
\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)
\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow2sinx.cosx=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)
c.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)
Đề là \(sin\left(5-\dfrac{\pi}{3}\right)\) hay \(sin\left(5x-\dfrac{\pi}{3}\right)\) nhỉ?
\(\left(5x-\dfrac{\pi}{3}\right)\) nha sorry nhầm tí