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+3t2−83t⇒4t2−83t+15=0 Discriminant (D)=(83)2−4(4)(15)=192−240=−48<0 So, the equation has imaginary roots, So, there are no solutions.