A. 2COS2X + SINX = SIN3X  SIN3X – SINX – 2COS2X = 0 2COS2XSINX – 2C...

Câu 2:a. 2cos2x + sinx = sin3x  sin3x – sinx – 2cos2x = 0 2cos2xsinx – 2cos2x = 0  cos2x = 0 hay sinx = 1   hay x = 22 k (k  Z) x = 4 k2b. log

2

(2x).log

3

(3x) > 1, đk 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 16 hay x > 1 log

2

x < -log

2

6 hay log

2

x > 0  0 < x <

3

x dx



, đặt u = x1  u

²

= x + 1  2udu = dxx