giải phương trình lượng giác

2

2ku

:)
[TEX]2cos^2(\frac{\pi}{2}.cos^2x) = 1 + cos(\pi sin2x)[/TEX]
[TEX]cos(\pi.cos^2x) = cos(\pi.sin2x)[/TEX]
[TEX]\pi cos^2x = \pm \pi sin2x + 2k\pi[/TEX]
[TEX]cos^2x = \pm sin2x + 2k[/TEX]
[TEX]1 + cos2x = 2sin2x + 4k[/TEX]
[TEX]cos2x - 2sin2x = 4k - 1[/TEX]
Để pt có nghiệm thì [TEX]1^2 + 2^2\geq(4k-1)^2[/TEX]
\Leftrightarrow[TEX]\frac{1- \sqrt{5}}{4} \leq k \leq \frac{1+\sqrt{5}}{4}, k \epsilon Z[/TEX]
\Rightarrow k = 0
\Rightarrow [TEX]cos^2x = \pm sin2x[/TEX]
[TEX]cosx(cosx \pm 2sinx) = 0[/TEX]
[TEX]\left[\begin{cosx = 0}\\{2sinx = \pm cosx}[/TEX]
[TEX]\left[\begin{cosx = 0}\\{tanx = \pm \frac{1}{2}[/TEX]
 
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