Sin2x+sin5x-sinx / cosx + cos2x + cos5x = tq2 доказать равенство

sin x + 1 2 sin 2 x = cos x + cos 2 x ⇔ ⇔ sin x + sin x cos x = cos x + 2 cos 2 x - 1 ⇔ ⇔ sin x (1 + cos x) = 1 + cos x + 2 (cos x - 1) (cos x + 1) ⇔ ⇔ 2 sin x cos 2 x 2 = (1 + cos x) (2 cos x - 1) ⇔ ⇔ sin x cos 2 x 2 = (2 cos x - 1) cos 2 x 2 ⇔ ⇔ (sin x - 2 cos x + 1) cos 2 x 2 = 0. sin⁡x+12sin⁡2x=cos⁡x+cos⁡2x ⇔⇔ sin⁡x+sin⁡xcos⁡x=cos⁡x+2cos2⁡x-1 ⇔⇔ sin⁡x (1+cos⁡x) = 1+cos⁡x+2 (cos⁡x-1) (cos⁡x+1) ⇔⇔ 2sin⁡xcos2⁡x2 = (1+cos⁡x) (2cos⁡x-1) ⇔⇔ sin⁡xcos2⁡x2 = (2cos⁡x-1) cos2⁡x2 ⇔⇔ (sin⁡x-2cos⁡x+1) cos2⁡x2=0.