4sinx + 2cosx = 2 + 3tanx (1)
đk: cosx # 0 <=> x # ± π/2 + k2π (k Є Z)
Đặt t = tan(x/2) --> sinx = 2t/(1 + t²) ; cosx = (1 - t²)/(1 + t²) ; tanx = 2t/(1 - t²)
--> (1) <=> 8t/(1 + t²) + 2(1 - t²)/(1 + t²) = 2 + 6t/(1 - t²)
<=> 8t(1 - t²) + 2(1 - t²)² = 2(1 - t²)(1 + t²) + 6t(1 + t²)
<=> 8t - 8t³ + 2 - 4t² + 2t^4 = 2 - 2t^4 + 6t + 6t³
<=> 4t^4 - 14t³ - 4t² + 2t = 0
<=> 2t(2t³ - 7t² - 2t + 1) = 0
<=> t = 0 ; 2t³ - 7t² - 2t + 1 = 0
** t = 0 <=> tan(x/2) = 0 <=> x/2 = kπ <=> x = 2kπ (k Є Z)
** 2t³ - 7t² - 2t + 1 = 0
<=> (2t + 1)(t² - 4t + 1) = 0
<=> 2t + 1 = 0 ; t² - 4t + 1 = 0
+) t = -1/2 --> tan(x/2) = -1/2 = tan(α)
--> x/2 = α + kπ --> x = 2α + kπ (k Є Z)
+) t² - 4t + 1 = 0 <=> t = 2 ± √3
. t = 2 + √3 --> tan(x/2) = 2 + √3 = tan(π/12)
--> x/2 = π/12 + kπ
--> x = π/6 + k2π (k Є Z)
. t = 2 - √3 --> tan(x/2) = 2 + √3 = tan(5π/12)
--> x/2 = 5π/12 + kπ
--> x = 5π/6 + k2π (k Є Z)