3 SIN 2 X  COS 2 X  2COS -1 X 2 3 SINXCOSX + 2COS2X = 2COSX...

Câu 2. 3 sin 2 x  cos 2 x  2cos -1 x

 2 3 sinxcosx + 2cos

²

x = 2cosx  ^{2 cos} x  3 s inx cos  x   1  0

 cosx = 0 hay 3 sinx + cosx = 1

1

 cosx = 0 hay 3

x _  _ 

2 sinx + 1

3 3

2 ^{cosx =}

2  cosx = 0 hay cos( ) cos

     hay 2

 x = 2

3 2

x    k  (k  Z).

2 k hay x k