c) sin5x + 2cosx + sin3x = 0
<=> 2sin4x.cosx + 2cosx = 0
<=> 2cosx(sin4x + 1) = 0
<=> cosx = 0 hay sin4x = -1
<=> x = π/2 + kπ hay 4x = - π/2 + k2π
<=> x = π/2 + kπ hay x = - π/8 + kπ/2
Vậy...

a) sin^2(2x) + cos2x + 1 = 0
<=> 1 - cos^2(2x) + cos2x + 1 = 0
<=> - cos^2(2x) + cos2x + 2 = 0
<=> cos2x = 2 (loại vì x ∈ [ - 1 ; 1])
hoặc cos2x = -1
<=> 2x = π + k2π ( k ∈ Z)
<=> x = π/2 + kπ( k ∈ Z)
Vậy...
b) sin3x + √3.cos3x = √2
Vì √( 1 + 3) = √4 = 2 > √2 nên 
 1/2.sin3x + √3/2.cos3x = √2/2
<=> cos π/3 + sin π/3 cos3x = 1/√2
<=> sin(3x + π/3) = sin(π/4)
<=> 3x + π/3 = π/4 + k2π ( k ∈ Z)
hoặc 3x + π/3 = 3π/4 + k2π( k ∈ Z)
<=> 3x = - π/12 + k2π ( k ∈ Z)
hoặc 3x = 5π/12 + k2π( k ∈ Z)
<=> x = - π/36 + k2π/3 ( k ∈ Z)
hoặc x = 5π/36 + k2π/3( k ∈ Z)
Vậy...