Đáp án:

1) $ x = \dfrac{π}{6} + k2π;  x = \dfrac{5π}{6} + k2π $

2) $ x = k\dfrac{π}{2}; x = (2k + 1)\dfrac{π}{20} $

3) $ x = ± \dfrac{π}{14} + k\dfrac{2π}{7}$

4) $ x = \dfrac{π}{12} + 2mπ; x = \dfrac{3π}{12} + (2n + 1)π$

6) $ x = k2π$

Giải thích các bước giải:

1) ĐKXĐ $ sin2x \neq 0; cos2x \neq 0 ⇔ x \neq k\dfrac{π}{4} (*)$

$ PT ⇔ 2sinxcos2x + cos2x = 1 (MSC : sin2xcos2x = 2sinxcosxcos2x)$

$ ⇔ 2sinxcos2x - 2sin²x = 0$

$ ⇔ 2sinx(cos2x - sinx) = 0 ( sinx \neq 0) $

$ ⇔ cos2x - sinx = 0 ⇔ 2sin²x + sinx - 1 = 0$

$ ⇔ (2sinx - 1)(sinx + 1) = 0 ⇔ 2sinx - 1 = 0$

$ ⇔ x = \dfrac{π}{6} + k2π;  x = \dfrac{5π}{6} + k2π ( TM (*))$

(vì $ sin2x \neq 0 ⇒ cosx \neq 0 ⇒ sinx \neq ± 1)$

2) ĐKXĐ $ : cos5x \neq 0 ⇔ x \neq \dfrac{π}{10} + k\dfrac{2π}{5} (*)$

$ PT ⇔ 2cos3x.sin5x = 2sin7xcos5x$

$ ⇔ sin8x + sin2x = sin12x + sin2x$

$ ⇔ sin12x = sin8x $

- TH1$: 12x = 8x + k2π ⇔ x = k\dfrac{π}{2} (TM(*))$

- TH1$: 12x = π - 8x + k2π ⇔ x = (2k + 1)\dfrac{π}{20} (TM(*))$

3)ĐKXĐ $: cos2x \neq 0 ⇔ x \neq \dfrac{π}{4} + kπ$

$ : cos5x \neq 0 ⇔ x \neq \dfrac{π}{10} + k\dfrac{2π}{5} (*)$

$ PT ⇔ sin2x.sin5x = cos2xcos5x$

$ ⇔ cos2xcos5x - sin2x.sin5x = 0$

$ ⇔ cos(2x + 5x) = 0 ⇔ cos7x = 0$

$ ⇔ 7x = ± \dfrac{π}{2} + k2π ⇔ x = ± \dfrac{π}{14} + k\dfrac{2π}{7}(TM(*))$

4) ĐKXĐ $ : 1 + 8sin2xcos²2x ≥ 0 (*)$

$PT ⇔ \left[ \begin{array}{l}4sin²(3x + \dfrac{π}{4}) = 1 + 8sin2xcos²2x \\sin(3x + \dfrac{π}{4}) ≥ 0\end{array} \right.$

$⇔ \left[ \begin{array}{l}2[1 - cos(6x + \dfrac{π}{2})] = 1 + 1 + 4sin4xcos2x \\ 2kπ ≤ 3x + \dfrac{π}{4} ≤ (2k + 1)π\end{array} \right.$

$⇔ \left[ \begin{array}{l}2(1 + sin6x) = 1 + 2(sin6x + sin2x) \\ 2kπ ≤ 3x + \dfrac{π}{4} ≤ (2k + 1)π\end{array} \right.$

$⇔ \left[ \begin{array}{l} sin2x = \dfrac{1}{2} (TM (*))\\ 2kπ ≤ 3x + \dfrac{π}{4} ≤ (2k + 1)π\end{array} \right.$

$⇔ \left[ \begin{array}{l} 2x = \dfrac{π}{6} + k2π; 2x = \dfrac{5π}{6} + k2π\\ 2kπ ≤ 3x + \dfrac{π}{4} ≤ (2k + 1)π\end{array} \right.$

$⇔ \left[ \begin{array}{l} x = \dfrac{π}{12} + kπ; x = \dfrac{5π}{12} + kπ\\ 2kπ ≤ 3x + \dfrac{π}{4} ≤ (2k + 1)π\end{array} \right.$

$⇔ \left[ \begin{array}{l} 3x = \dfrac{π}{4} + 3kπ; 3x = \dfrac{5π}{4} + 3kπ\\ 2kπ ≤ 3x + \dfrac{π}{4} ≤ (2k + 1)π\end{array} \right.$

$⇔ \left[ \begin{array}{l} 3x + \dfrac{π}{4} = \dfrac{π}{2} + 3kπ; 3x + \dfrac{π}{4} = \dfrac{3π}{2} + 3kπ\\ 2kπ ≤ 3x + \dfrac{π}{4} ≤ (2k + 1)π(**)\end{array} \right.$

Chọn $ 3x + \dfrac{π}{4} = \dfrac{π}{2} + 6mπ; 3x + \dfrac{π}{4} = \dfrac{3π}{2} + 3(2n + 1)π (TM (**))$

$ ⇔ 3x = \dfrac{π}{4} + 6mπ; 3x = \dfrac{3π}{4} + 3(2n + 1)π$

$ ⇔ x = \dfrac{π}{12} + 2mπ; x = \dfrac{3π}{12} + (2n + 1)π$

5) $ \sqrt{2} - [sin(2x + \dfrac{π}{4}) + cos(2x + \dfrac{π}{4})] + 2(2 - \sqrt{2})sinx - 4 = 0$

$ ⇔ \sqrt{2} - \sqrt{2}sin[(2x + \dfrac{π}{4}) + \dfrac{π}{4}] + 2(2 - \sqrt{2})sinx - 4 = 0$

$ ⇔ \sqrt{2} - \sqrt{2}sin(2x + \dfrac{π}{2}) + 2(2 - \sqrt{2})sinx - 4 + \sqrt{2} = 0$

$ ⇔ \sqrt{2}(1 - cos2x) + 2(2 - \sqrt{2})sinx - 4 = 0$

$ ⇔ \sqrt{2}sin²x + (2 - \sqrt{2})sinx - 2 = 0$

Đây là PT bậc 2 nhưng nghiệm xấu quá, bạn chịu khó giải tiếp

6) $PT ⇔ cos10x + 2cos²4x = cosx + 2cosx(4cos³3x - 3cos3x)$

$ ⇔ cos10x + (1 + cos8x) = cosx + 2cosxcos3(3x)$

$ ⇔ (cos10x + cos8x) + 1 = cosx + 2cosxcos9x$

$ ⇔ 2cosxcos9x + 1 = cosx + 2cosxcos9x$

$ ⇔ cosx = 1 ⇔ x = k2π$