B0.75Đ’(1,5Đ’) 2− = − ⇔ −Π = −3 SIN 2X COS2X 2 SIN(2X )6 2Π Π Π − =...
1.b0.75đ’(1,5đ’) 2− = − ⇔ −π = −3 sin 2x cos2x 2 sin(2x )6 2π π π − = − + π = − + π2x k2 x k π π6 4 24⇔ − = − ⇔ ⇔ ∈sin(2x ) sin( ) (k Z)π1756 4 − = + π = + πx k2x k2 4 4 24