The number of all the possible triplets (a1,a2,a3) such that a1+a2cos(2x)+a3sin2(x)=0 for all x is
infinite
Since a1+a2cos2x+a3sin2x=0firakkx
Putting x=0andx=π2, we get
a1+a2=0anda1−a2+a3=0⇒a2=−a1anda3=−2a1
Therefore, the given equation becomes
a1−a1cos2x−2a1sin2x=0,∀x
or a1(1−cos2x−2sin2x)=0,∀x
or a1(2sin2x−2sin2x)=0,∀x
The above is satisfied for all values of a1.
Hence, the infinite number of triplets (a1,−a1,−2a1) is possible.
Alternative Method:
a1+a2cos2x+a3sin2x=0forrealx∴a1+a2(1−2sin2x)+a3sin2x=0forrealx
∴(a1+a2)+(a3−2a2)sin2x=0 for real x
∴a1+a2=0anda3−2a2=0∴a2=−a1anda3=2a2=−2a1
hence, infinite number of triplets (a1,−a1,−2a1) exist.