Given equation is, √3sinx+cosx=4 Let sinx=t ∵cosx=√1−sin2x∴cosx=√1−t2 Now, √3t+√1−t2=4,√1−t2=4−√3t Squaring both sides, we get 1−t2=16+3t2−8√3t ⇒4t2−8√3t+15=0 Discriminant (D)=(8√3)2−4(4)15 =192−240=−48<0 So, the equation has imaginary roots, So, there are no solutions.