Tentukan himpunan penyelesaiannya pada interval 0 ≤ x ≤ 2π a. 2 cos²x = cos x b. tan x = 1/tan x c. √2 (cos x + sin x) = 2

a.
2 cos²x - cos x = 0
cos x [2 cos x  -  1] = 0
cos x = 0                     cos x = 1/2
x = π/2 , 3π/2               x = π/6 , 11π/6

b.
tan x = 1/tan x  (x ≠ {0, π, 2π})
tan²x = 1
tan x = 1
x = π/4 , 5π/4

c. 
√2 (cos x + sin x) = 2
cos x + sin x = √2
cos²x + 2 sinx . cosx + sin²x = 2
1 + 2 tanx + tan²x = 2/cos²x
tan²x - 2 sec²x + 2 tanx + 1 = 0
tan²x - 2 - 2 tan²x + 2 tanx + 1 = 0
-tan²x + 2 tanx - 1 = 0
tan²x - 2 tanx + 1 = 0
(tanx  -  1)² = 0
tanx = 1
x = π/4 , 5π/4