1 bài PT ai giải giúp tớ

T

thong1990nd

giải phương trình sau
[tex](cosx+1)(cos2x+1)(cos3x+1)=\frac {1}{2} [/tex]
ha bac
\Leftrightarrow [TEX](cos(\frac{x}{2}).cosx.cos(\frac{3x}{2}))^2=\frac{1}{16}[/TEX]
\Leftrightarrow [TEX]\left[\begin{cos(\frac{x}{2}).cosx.cos(\frac{3x}{2})=1/4}\\{ cos(\frac{x}{2}).cosx.cos(3x/2)= \frac{-1}{4}}[/TEX]
(1) \Leftrightarrow [TEX](cosx+cos2x)cosx=\frac{1}{2}[/TEX]
\Leftrightarrow [TEX](cosx+2cos^2x-1)cosx=\frac{1}{2}[/TEX]
\Leftrightarrow [TEX]2cos^3x+cos^2x-cosx-\frac{1}{2}=0[/TEX]
nham [TEX]cosx=\frac{1}{\sqrt[]{2}}[/TEX] la 1 no sau do chia la ra
(2) ttu nham [TEX]cosx=+-\frac{1}{2}[/TEX] :D
 
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