A. 2COS2X + SINX = SIN3X ⇔ SIN3X – SINX – 2COS2X = 0 ⇔ 2COS2XSINX...

Câu 2: a. 2cos2x + sinx = sin3x ⇔ sin3x – sinx – 2cos2x = 0 ⇔ 2cos2xsinx – 2cos2x = 0 ⇔ cos2x = 0 hay sinx = 1 π ₊ ππ + π (k ∈ Z) ⇔ x = hay x = 24 k 22 kb. log

2

(2x).log

3

(3x) > 1, ñk x > 0 ⇔ log

3

x + log

₂

x + log

₂

x.log

₃

x > 0 ⇔ log

3

2(log

₂

x)

²

+ (log

₃

2 + 1)log

₂

x > 0 ⇔ log

2

x < -log

2

6 hay log

2

x > 0 ⇔ 0 < x < 16 hay x > 1

3

x dx

∫

^{, ñặt u = x+1 ⇒ u}

²

= x + 1 ⇒ 2udu = dx