Question

[Phạm Dương Ngọc Nhi]

\(1,sin^{2008}x+cos^{2008}x=1\)

\(2,sin^5x+cos^5x+sin2x+cos2x=1+\sqrt{2}\)

\(3,4cos^2x+3tan^2x-4\sqrt{3}cosx+2\sqrt{3}tanx+4=0\)

Answer 1

[Nguyễn Việt Lâm]

3.

\(\Leftrightarrow\left(4cos^2x-4\sqrt{3}cosx+3\right)+\left(3tan^2x+2\sqrt{3}tanx+1\right)=0\)

\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)^2+\left(\sqrt{3}tanx+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2cosx-\sqrt{3}=0\\\sqrt{3}tanx+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\tanx=-\frac{1}{\sqrt{3}}\end{matrix}\right.\)

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

\(\Rightarrow x=-\frac{\pi}{6}+k2\pi\)

Answer 2

[Nguyễn Việt Lâm]

2.

Do \(-1\le cosx;sinx\le1\Rightarrow\left\{{}\begin{matrix}sin^5x\le sin^2x\\cos^5x\le cos^2x\end{matrix}\right.\)

\(\Rightarrow sin^5x+cos^5x\le sin^2x+cos^2x=1\)

Lại có: \(sin2x+cos2x=\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)\le\sqrt{2}\)

\(\Rightarrow sin^5x+cos^5x+sin2x+cos2x\le1+\sqrt{2}\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}sinx=1\\sin2x+cos2x=\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}cosx=1\\sin2x+cos2x=\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}sinx=1\\-1=\sqrt{2}\left(vn\right)\end{matrix}\right.\\\left\{{}\begin{matrix}cosx=1\\2cos^2x-1=\sqrt{2}\left(vn\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm