Question

\(1.\left(sinx+cosx\right)^3+sinxcosx-1=0\)

\(2.\left(sinx+cosx\right)^4-3sin2x-1=0\)

\(3.sin^3x+cos^3x+2\left(sinx+cosx\right)-3sin2x=0\)

\(4.\left(sinx-cosx\right)^3=1+sinxcosx\)

5.\(sinx+cosx+2+tanx+cotx+\frac{1}{sinx}+\frac{1}{cosx}=0\)

Answer 1

1.

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

Pt trở thành:

\(t^3+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow2t^3+t^2-3=0\)

\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)

\(\Leftrightarrow t=1\)

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

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

\(\Leftrightarrow...\)

Answer 2

b.

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=t^2-1\end{matrix}\right.\)

Pt trở thành:

\(t^4-3\left(t^2-1\right)-1=0\)

\(\Leftrightarrow t^4-3t^2+2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}1+sin2x=1\\1+sin2x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow...\)

Answer 3

3.

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)+2\left(sinx+cosx\right)-6sinx.cosx=0\)

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

Pt trở thành:

\(t\left(1-\frac{t^2-1}{2}\right)+2t-3\left(t^2-1\right)=0\)

\(\Leftrightarrow-t^3-6t^2+7t+6=0\)

Nghiệm của pt bậc 3 này rất xấu, chắc bạn ghi ko đúng đề bài

Answer 4

4.

Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{1-t^2}{2}\end{matrix}\right.\)

Pt trở thành:

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

\(\Leftrightarrow2t^3+t^2-3=0\)

\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)

\(\Leftrightarrow t=1\)

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

\(\Leftrightarrow...\)

Answer 5

5.

ĐKXĐ:

\(sinx+cosx+2+\frac{sinx}{cosx}+\frac{cosx}{sinx}+\frac{sinx+cosx}{sinx.cosx}=0\)

\(\Leftrightarrow sinx+cosx+\frac{\left(sinx+cosx\right)^2}{sinx.cosx}+\frac{sinx+cosx}{sinx.cosx}=0\)

\(\Leftrightarrow\left(sinx+cosx\right)\left(1+\frac{sinx+cosx}{sinx.cosx}+\frac{1}{sinx.cosx}\right)=0\)

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

Xét (1):

\(\Leftrightarrow sinx+cosx+sinx.cosx+1=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)+2sinx.cosx+1+1=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)+\left(sinx+cosx\right)^2+1=0\)

\(\Leftrightarrow\left(sinx+cosx+1\right)^2=0\)

\(\Leftrightarrow sinx+cosx=-1\)

\(\Leftrightarrow...\)