sin^4xcos^x2x-cos^4xsin^2x=cos2x

sin^4xcos^x2x-cos^4xsin^2x=cos2x
- Предмет:
  Алгебра
- Автор:
  fermínghv5
- 6 лет назад

Ответы 1

Видимо, уравнение таково: $\sin^4x \cos^2x-\cos^4x \sin^2x=\cos2x$
$\sin^2x \cos^2x( \sin^2x- \cos^2x) = \cos2x \\\frac{1}{4} \sin^22x \cdot (-\cos2x )= \cos2x \\\cos2x ( 1+\frac{1}{4} \sin^22x)= 0$
$\cos2x = 0$    или    $1+\frac{1}{4} \sin^22x = 0$
$2x= \frac{\pi}{2} + \pi k$    $\sin^22x =-4$
$x= \frac{\pi}{4} +\frac{\pi k}{2} ,\ k \in Z$     решений нет
Ответ: $\frac{\pi}{4} +\frac{\pi k}{2} ,\ k \in Z$
- Автор:
  catherinemason
- 6 лет назад
- 0
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sin^4xcos^x2x-cos^4xsin^2x=cos2x

Ответы 1

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