If cos4⁡θ+α and sin4⁡θ+α are the roots of the equation x2+2bx+b=0 and cos2⁡θ+β,sin2⁡θ+β are the roots of the equation  x2+4x+2=0, then values of b are

We have, cos2⁡θ−sin2⁡θ=cos⁡2θ⇒ cos4⁡θ−sin4⁡θ=cos⁡2θ⇒ (−2b)2−4b=(−4)2−4×2(since L.H.S. is difference of roots of first equation and R.H.S. is difference of roots of second equation)or   4b2−4b=16−8=8or   4b2−4b−8=0or   b2−b−2=0or   (b+1)(b−2)=0or   b=2,−1