Question

1)log6(48)-log6(4)+log6(3) 2)cos(pi-x)+sin(3pi2-x)+2cos(2pi-x) , ! =)

Answer 1

1) $log_{6}48-log_{6}4+log_{6}3=log_{6}( \frac{48*3}{4})=log_{6}36=2$

2) $cos( \pi -x)+sin( \frac{3 \pi }{2}-x)+2cos(2 \pi -x)=-cosx-cosx+2cosx=0$