Đáp án:
Giải thích các bước giải:
1) $PT ⇔ 2sin²4x - 2sin²2x + 2sin²3x - 2sin²x = 0$
$ ⇔ (1 - cos8x) - (1 - cos4x) + (1 - cos6x) + (1 - cos2x) = 0$
$ ⇔ (cos4x - cos8x) + (cos2x - cos6x) = 0$
$ ⇔ 2sin6xsin2x + 2sin4xsin2x = 0$
$ ⇔ 2sin2x(sin6x + sin4x) = 0 $
$ ⇔ 4sin2xsin5xcosx = 0$
@ $ sin2x = 0 ⇔ 2x = kπ ⇔ x = \dfrac{kπ}{2}$
@ $ sin5x = 0 ⇔ 5x = kπ ⇔ x = \dfrac{kπ}{5}$
2) $ x ∈ (0; π) ⇒ 0 < sinx < 1$
$ ⇒ \sqrt{1 - cos2x} = \sqrt{2sin²x} = \sqrt{2}|sinx| = \sqrt{2}sinx$
$ PT ⇔ \dfrac{2cos2xsinx}{\sqrt{2}sinx} = \sqrt{2}cos(2x - \dfrac{π}{4})$
$ ⇔ cos2x = cos(2x - \dfrac{π}{4}) ⇔ 2x = - (2x - \dfrac{π}{4}) + k2π $
$ ⇔ 4x = \dfrac{π}{4} + k2π ⇔ x = \dfrac{π}{16} + \dfrac{kπ}{2}$
Do $ 0 < x < π ⇔ 0 < \dfrac{π}{16} + \dfrac{kπ}{2} < π$
$ ⇔ - \dfrac{π}{16} < \dfrac{kπ}{2} < \dfrac{15π}{16} $
$ ⇔ - \dfrac{1}{8} < k < \dfrac{15}{8} ⇒ k = 0; k = 1$
$ ⇒ x = \dfrac{π}{16}; x = \dfrac{9π}{16}$
