[TEX]\ 5(sinx+\frac{cos3x+sin3x}{1+2sin2x})= 3 + cos2x [/TEX](1)
Đk:[TEX]\ sin2x \neq \frac{-1}{2}[/TEX]
(1) \Leftrightarrow [TEX]\ 5.\frac{sinx + 2 sinxsin2x + cos3x + sin3x}{1+2sin2x} = 3 + cos2x[/TEX]
\Leftrightarrow [TEX]\ 5.\frac{sinx + (cosx - cos3x) + cos3x + sin3x}{1+2sin2x} = 3 + cos2x [/TEX]
\Leftrightarrow [TEX]\ 5\frac{2sin2xcosx + cosx}{1 + 2sin2x} = 3 + cos2x[/TEX]
\Leftrightarrow [TEX]\ 3cosx = 3 + 2 cos^{2}x - 1[/TEX]
\Leftrightarrow [TEX]\ 2cos^{2} - 5cosx + 2 = 0[/TEX]
Đặt t = cosx (|t| \leq 1)
\Rightarrow Pttt: [TEX]\2t^{2} - 5t + 2 =0[/TEX]
\Leftrightarrow [TEX]\ t = 2 (L) (or) t = \frac{1}{2}[/TEX]
\Rightarrow [TEX]\ cosx = \frac{1}{2}[/TEX]
\Leftrightarrow [TEX]\ x = \frac{\pm \pi }{3} + k2\pi [/TEX]
- Với [TEX]\ x = \frac{ \pi }{3} + k2\pi[/TEX]
Vì [TEX]\ x \in (0;2\pi)[/TEX] \Rightarrow [TEX]\ 0< \frac{\pi}{3} + k2\pi < 2\pi [/TEX]
\Leftrightarrow [TEX]\ \frac{-1}{6} < k < \frac{5}{6}[/TEX]
Vì [TEX]\ k \in Z[/TEX] \Rightarrow [TEX]\ k = 0 \Rightarrow x=\frac{\pi}{3}[/TEX]
- Với [TEX]\ x = \frac{ - \pi }{3} + k2\pi [/TEX]
Vì [TEX]\ x \in (0;2\pi)[/TEX] \Rightarrow [TEX]\ 0< \frac{-\pi}{3} + k2\pi < 2\pi [/TEX]
\Leftrightarrow [TEX]\ \frac{1}{6} < k < \frac{7}{6}[/TEX]
Vì [TEX]\ k \in Z[/TEX] \Rightarrow k = 1 \Rightarrow[TEX]\ x=\frac{7\pi}{3}[/TEX]
Vậy pt đã cho có 2 nghiệm trong [TEX]\(0;2\pi): x=\frac{\pi}{3}; x=\frac{7\pi}{3}[/TEX]
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