Đáp án:
Giải thích các bước giải:
1. $sin(10x + \dfrac{17π}{2}) = sin(10x + \dfrac{π}{2} + 8π)$
$ = sin(10x + \dfrac{π}{2}) = cos10x$
$ PT ⇔ 2sin²2x - 2cos²8x = 2cos10x$
$ ⇔ (1 - cos4x) - (1 + cos16x) = 2cos10x$
$ ⇔ (cos16x + cos4x) + 2cos10x = 0$
$ ⇔ 2cos10xcos6x + 2cos10x = 0$
$ ⇔ 2cos10x(cos6x + 1) = 0$
- TH1 $: cos10x = 0 ⇔ 10x = (2k + 1)\dfrac{π}{2} ⇔ x = (2k + 1)\dfrac{π}{20} $
- TH2 $: cos6x = - 1 ⇔ 6x = (2k +1)π ⇔ x = (2k + 1)\dfrac{π}{6} $
2. Tương tự câu 1$: sin(2x + \dfrac{17π}{2}) = cos2x$
$ cos(x - \dfrac{15π}{2}) = cos(x + \dfrac{π}{2} - 8π)$
$ = cos(x + \dfrac{π}{2}) = - sinx$
$ PT ⇔ cos2x + 3sinx = 1 + 2sinx$
$ ⇔ (1 - cos2x) - sinx = 0$
$ ⇔ 2sin²x - sinx = 0$
$ ⇔ sinx(2sinx - 1) = 0$
- TH1 $: sinx = 0 ⇔ x = kπ$
- TH2 $: sinx = \dfrac{1}{2} ⇔ x = \dfrac{π}{6} + 2kπ; x = \dfrac{5π}{6} + 2kπ $
3. $PT ⇔ - sinx(1 - 2sin²x) - cosx(2cos²x - 1) - cos2x = 0$
$ ⇔ - sinxcos2x - cosxcos2x - cos2x = 0$
$ ⇔ - cos2x(sinx + cosx + 1) = 0$
- TH1 $: cos2x = 0 ⇔ 2x = (2k + 1)\dfrac{π}{2} ⇔ x = (2k + 1)\dfrac{π}{4} $
- TH2 $: sinx + cosx = - 1 $
$ ⇔ \sqrt{2}sin(x + \dfrac{π}{4}) = - 1 ⇔ sin(x + \dfrac{π}{4}) = - \dfrac{\sqrt{2}}{2}$
$ x + \dfrac{π}{4} = - \dfrac{π}{4} + k2π ⇔ x = - \dfrac{π}{2} + k2π $
$ x + \dfrac{π}{4} = - \dfrac{3π}{4} + k2π ⇔ x = - π + k2π $
