Solve the equation sin^8 x − cos^4 2x + cos^8 x = 0 .

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pramodpandey | College Teacher | (Level 3) Valedictorian

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We have given

sin^8 (x)-cos^4(2x)+cos^8(x)=0

LHS.=sin^8 (x)-cos^4(2x)+cos^8(x)

=sin^8(x)-(cos^2(x)-sin^2(x))^4+cos^8(x)

=sin^8(x)-{cos^8(x)-4cos^6(x)sin^2(x)+6cos^4(x)sin^4(x)

-4cos^2(x)sin^6(x)+sin^8(x)}+cos^8(x)

=4cos^6(x)sin^2(x)-6cos^4(x)sin^4(x)+4cos^2(x)sin^6(x)

=2cos^2(x)sin^2(x){2cos^4(x)-3cos^2(x)sin^2(x)+2sin^4(x)}

=2cos^2(x)sin^2(x){2((cos^2(x)+sin^2(x))^2-2sin^2(x)cos^2(x))-3sin^2(x)cos^2(x)}

=2cos^2(x)sin^2(x){2-7sin^2(x)cos^2(x)}

2cos^2(x)sin^2(x){2-7sin^2(x)cos^2(x)}=0

either

(2cos(x)sin(x))^2=0

sin(2x)=0

2x=n pi

x=(n/2).pi

or

sin^2(x)cos^2(x)=2/7

4sin^2(x)cos^2(x)=8/7

(sin(2x))^2=8/7

sin(2x)=+-sqrt(8/7)

It is not possible because  -1<=sin(2x) <=1

Thus

x=(n/2)pi

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