1/ [TEX]2cos^2 \frac{3x}{5}+1=3cos(\frac{4x}{5})[/TEX]
2/ [TEX]cos3x + \sqrt[2]{2-cos^2 3x}=2(1+sin^2 2x)[/TEX]
1.
\Leftrightarrow[TEX]1+cos.\frac{6x}{5}+1=3cos.\frac{4x}{5}[/TEX]
\Leftrightarrow[TEX]4cos^3\frac{2x}{5}-3cos\frac{2x}{5}+2=6cos^2\frac{2x}{5}-3[/TEX]
2.
áp dụng BĐT bunhia ta có \RightarrowVT\leq2
VP\geq2
dấu "=" xảy ra \Leftrightarrow[TEX]\left{\begin{cos3x=\sqrt{2-cos^23x}}\\{sin2x=1} [/TEX]
\Leftrightarrow[TEX]\left{\begin{cos^23x=1}\\{sin2x=0} [/TEX]
\Leftrightarrow[TEX]\left{\begin{sin3x=0}\\{sin2x=0} [/TEX]
\Rightarrow[TEX]x=k2\pi,k \in Z[/TEX]