bài 4
+  1 + cos (x) + cos (2x) = 0 
1 + cos (x) + cos (x + x) = 0 
1 cos (x) + cos (x) ^ 2 - sin (x) ^ 2 = 0 
1 + cos (x) + cos (x) ^ 2 - (1 - cos (x) ^ 2)) = 0 
1 + cos (x) + 2cos (x) ^ 2 - 1 = 0 
cos ) + 2cos (x) ^ 2 = 0 
cos (x) (1 + 2cos (x)) = 0 
=> cos (x) = 0, x = π / 2, 3π / 2 
=> 1 + 2cos (x) = 0, cos (x) = -1 / 2, x = 2π / 3, 4π / 3 

Vậy x = π / 2, 2π / 3, 4π / 3, 3π / 2.
4  3cosx-2sin2x = 0: 
3cosx-4sinxcosx 0 = 
cosx (3-4sinx) = 0 
cosx = 0 hoặc 3-4sinx = 0 
x = π / 2 + πk, k∈Z hoặc sinx = 3/4 x = arcsin (3/4) + 2πn, n∈Z; x = π-arcsin (3/4) + 2πm, m∈Z. b) 2cos2x = 1-sinx; 2 (1-sin²x) = 1-sinx; 2sin²x-sinx-1 = 0; D = 1 + 8 = 9 sinx = (1-3) / 4 = -1 / 2 hoặc sinx = (1 + 3) / 4 = 1 x = (- π / 6) + 2πk, k∈Z hoặc x = π - (- π / 6) + 2πn, n∈Z; x = π / 2 + 2πm, m∈Z On vào TA e) pi / 2 + πk; arcsin (3/4) + 2πn; π-arcsin (3/4) + 2πm ; k, n, m∈Z. b) (- π / 6) + 2πk; (7π / 6) + 2πn; π / 2 + 2πm; k, n, m∈Z