Chủ đề này có 1 trả lời

[nbngoc95]

GPT: $4cos^{2}2x+2cos2x-1=\frac{1}{2cos5x}$

[Mylovemath]

GPT: $4cos^{2}2x+2cos2x-1=\frac{1}{2cos5x}$

PT $ \Leftrightarrow 2(cos4x +1) + 2cos2x - 1 = \frac{1}{2cos5x}$

$ \Leftrightarrow 2cos4x + 2 + 2cos2x -1 = \frac{1}{2cos5x}$

$\Leftrightarrow 2cos4x + 2cos2x + 1 = \frac{1}{2cos5x}$

$\Leftrightarrow 4cos4xcos5x + 4cos2xcos5x + 2cos5x = 1$ (ĐK: $x \neq \frac{\pi}{10} + \frac{k\pi}{5}$)

$\Leftrightarrow 2cos9x + 2cosx + 2cos7x + 2cos3x + 2cos5x =1$

$\Leftrightarrow cosx + cos3x + cos5x + cos7x + cos9x = \frac{1}{2}$

$\Leftrightarrow 2sinxcosx + 2sinxcos3x + 2sinxcos5x + 2sinxcos7x + 2sinxcos9x = sinx$

$\Leftrightarrow sin2x - sin2x + sin4x - sin4x + sin6x - sin6x + sin8x - sin8x + sin10x = sinx$

$\Leftrightarrow sin10x = sinx$

$\Leftrightarrow \Leftrightarrow \left[ \begin{array}{l} 10x = x + k2\pi\\ 10x = \pi - x + k2\pi \end{array} \right.$

$\Leftrightarrow \Leftrightarrow \left[ \begin{array}{l} x = \frac{k2\pi}{9}\\ x = \frac{\pi + k2\pi}{9} \end{array} \right.$

Vậy : phương trình có nghiệm (TMĐK)