.if x+270[cos2Π/7+cos4Π/7+cos6Π/7=3746,then find the value of x.

Multiply both sides with sin(π7) hence we have that

sin(π7)cos(2π7)+sin(π7)cos(4π7)+sin(π7)⋅cos(6π7)=−12⋅sin(π7)
Then because sinA⋅cosB=12⋅[sin(A−B)+sin(A+B)]we rewrite this as

(12)[sin(π7−2π7)+sin(π7+2π7)]+(12)[sin(π7−4π7)+sin(π7+4π7)]+(12)[sin(π7−6π7)+sin(π7+6π7)]=−sin(π7)2

We'll divide by (12) both sides: 
−sin(π7)+sin(3π7)−sin(3π7)+sin(5π7)−sin(5π7)+sin(7π7)=−sin(π7)

Cancelling the same terms yields to 
−sin(π7)+sin(π)=−sin(π7)

But sin(π)=0 hence

−sin(π7)=−sin(π7)

Since we have the same results in both sides, the given identity is true.