Question

2(sin^6x+cos^6x)+t(sin^4x+cos^4x)=-1 ,then t equal to the​

Answer 1

Answer:

Step-by-step explanation:

sin^2A+cos^2A=1

2)1-sin^2A=cos^2A

3)1-cos^2A=sin^2A

=2(sin^6A+cos^6A)-3(sin^4A+cos^4A)+1

=2sin^6A+2cos^6A-3sin^4A-3cos^4A

+sin^2A+cos^2A

=2sin^6A+2cos^6A-2sin^4A-2cos^4A-sin^4A-cos^4A+sin^2A+cos^2A

=2sin^6A-2sin^4A+2cos^6A-2cos^4A

+sin^2A-sin^4A+cos^2A-cos^4A

=-2sin^4A(1-sin^2A)-2cos^4A(1-cos^2A)

sin^2A(1-sin^2A)+cos^2A(1-cos^2A)

=-2sin^4Acos^2A-2cos^4Asin^2A

+sin^2Acos^2A+cos^2Asin^2A

=-2sin^2Acos^2A(sin^2A+cos^2A)

+2sin^2Acos^2A

=-2sin^2Acos^2A+2sin^2Acos^2A

=0

Hence,

2(sin^6A+cos^6A)-3(sin^4A+cos^4A)+1=0

there by t=3 ,on comparing