3sinx+4cosx=5
=> 3/5 sinx +4/5 cosx = 1
let cosA=3/5 => sinA=4/5
=> cosAsinx + sinAcosx = 1
=> sin(x+A) = 1
Now,
4sinx - 3cosx
= 5(4/5sinx - 3/5 cosx) [multipying numerator and denominator by 5]
= 5(sinAsinx - cosAcosx)
= -5{cos(x+A)} = -5[root{1-(sin(x+A)^2)}] = -5 x 0 = 0 Ans 0
TRICK: if asinx+bcosx is given then multiply numerator and denominator by root(a^2 +b^2)
this method is useful in many questions