Sinx+cos2x+sin3x+cos4x+0

sin x + sin 3x = 2 cos x sin 2x cos 2x + cos 4x = 2 cos x cos 3x 2 cos x (sin 2x + cos 3x) = 0 cos x = 0:  x = (2n + 1) pi/2; n ∈ Z sin 2x + cos 3x = 0 cos 3x = cos (2x + x) = cos 2x cos x - sin 2x sin x = cos 2x cos x - 2 sin^2 x cos x 2 sin x cos x + cos 2x cos x - 2 sin^2 x cos x = 0  | : cos x 2 sin x + 1 - 2 sin^2 x - 2 sin^2 x = 0 4 sin^2 x - 2 sin x - 1 = 0 sin x = t; 4t^2 - 2t - 1 = 0 D/4 = 1 + 4 = 5 t = (1 +- sqrt(5))/4 sin x = (1 +- sqrt(5))/4 Ответ. [latex]x=\dfrac\pi2(2n+1),\quad n\in\mathbb Z\\ x=(-1)^m\arcsin\dfrac{1-\sqrt5}{4}+\pi m,\quad m\in\mathbb Z\\ x=(-1)^k\arcsin\dfrac{1+\sqrt5}4+\pi k,\quad k\in\mathbb Z[/latex]