Question

[Ngân Lại]

Giải các phương trình sau:
1) 1 + tanx = 2sinx + \(\frac{1}{cosx}\)

2) 1 + sin³2x + cos³2x = \(\frac{3}{2}\)sin4x

Answer 1

[Nguyễn Việt Lâm]

2/

\(\Leftrightarrow1+\left(sin2x+cos2x\right)^3-3sin2x.cos2x\left(sin2x+cos2x\right)=3sin2x.cos2x\)

Đặt \(sin2x+cos2x=t\Rightarrow\left|t\right|\le\sqrt{2}\)

\(t^2=1+2sin2x.cos2x\Rightarrow sin2x.cos2x=\frac{t^2-1}{2}\)

Pt trở thành:

\(1+t^3-\frac{3}{2}\left(t^2-1\right).t=\frac{3}{2}\left(t^2-1\right)\)

\(\Leftrightarrow t^3+3t^2-3t-5=0\)

\(\Leftrightarrow\left(t+1\right)\left(t^2+2t-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=-1+\sqrt{6}\left(l\right)\\t=-1-\sqrt{6}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin2x+cos2x=-1\)

\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=-1\)

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

\(\Rightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

1/

ĐKXĐ: ...

\(\Leftrightarrow cosx+sinx=2sinx.cosx+1\)

\(\Leftrightarrow sinx+cosx=2sinx.cosx+sin^2x+cos^2x\)

\(\Leftrightarrow sinx+cosx=\left(sinx+cosx\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}sinx+cosx=0\\sinx+cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=0\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=k\pi\\x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k2\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)