The equation `sin^4 x + cos^4 x + sin2x + alpha = 0` is solvable for

Options

A. `alpha in [(-1)/(2),(1)/(2)]`
B. `alpha in [-3,1]`
C. `alpha in [(-3)/(2),(1)/(2)]`
D. `alpha in [-1,1]`

Answer

B, C

Explanation

`cosB+cosC=4sin^(2)(A//2)`
`implies 2cos.(B+C)/(2)cos.(B-C)/(2)=4sin^(2).(A)/(2)`
`implies2cos((pi)/(2)-(A)/(2))cos.(B-C)/(2)=4sin.(A)/(2)sin((pi)/(2)-(B+C)/(2))`
`cos((B-C)/(2))=2cos((B+C)/(2))" "[becausesin.(A)/(2)ne0]`
Multiplying with `sin((B+C)/(2))` on both sides
`cos((B+C)/(2))sin((B+C)/(2))=sin(B+C)`
`(1)/(2)(sinB+sinC) =sinAimpliesb+c=2a`
`implies AB+AC=2a`
Also `AB+AC gt BC=a`
So locus of A is an ellipse with focii at B and C and the length of the major axis is 2a.