Question

sinx+cos2x+sin3x+cos4x+0

Answer 1

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. $x=dfracpi2(2n+1),quad ninmathbb Z\ x=(-1)^marcsindfrac{1-sqrt5}{4}+pi m,quad minmathbb Z\ x=(-1)^karcsindfrac{1+sqrt5}4+pi k,quad kinmathbb Z$