20,sin^(4)x

The number of distinct solutions of 5/4cos^(2)2x+cos^(4)x+sin^(4)x+cos^(6)x+sin^(6)x=2 is [0,20pi]

Solev (sin^(2) 2x+4 sin^(4) x-4 sin^(2) x cos^(2) x)/(4-sin^(2) 2x-4 sin^(2) x)=1/9 .

Solve (sin^(2) 2x+4 sin^(4) x-4 sin^(2) x cos^(2) x)/(4-sin^(2) 2x-4 sin^(2) x)=1/9 .

(d^(20))/(dx^(20))(2cos x cos3x)=2^(20)(cos2x-2^(20)cos4x)(b)2^(20)(cos2x+2^(20)cos4x)(c)2^(20)(sin2x+2^(20)sin4x)(d)2^(20)(sin2x-2^(20)sin4x)

(d^(20))/(dx^(20))(2cosxcos3x)= 2^(20)(cos2x-2^(20)cos4x) (b) 2^(20)(cos2x+2^(20)cos4x) (c) 2^(20)(sin2x+2^(20)sin4x) (d) 2^(20)(sin2x-2^(20)sin4x)

(d^(20)y)/(dx^(20))(2cosxcos3x)i se q u a l to a)2^(20)(cos2x-2^(20)os3x) b)2^(20)(cos2x+2^(20)cos4x) c)2^(20)(sin2x+2^(20)sin4x) d)2^(20)(sin2x-2^(20)sin4x)

(d^(20)y)/(dx^(20))(2cosxcos3x)is equal to 2^(20)(cos2x-2^(20)os3x) 2^(20)(cos2x+2^(20)cos4x) 2^(20)(sin2x+2^(20)sin4x) 2^(20)(sin2x-2^(20)sin4x)

(d^(20))/(dx^(20))(2cosxcos3x) is equal to (a) 2^(20)(cos2x-2^(20)cos3x) (b) 2^(20)(cos2x+2^(20)cos4x) (c) 2^(20)(sin2x+2^(20)sin4x) (d) 2^(20)(sin2x-2^(20)sin4x)

(d^(20))/(dx^(20))(2cosxcos3x) is equal to (a) 2^(20)(cos2x-2^(20)cos3x) (b) 2^(20)(cos2x+2^(20)cos4x) (c) 2^(20)(sin2x+2^(20)sin4x) (d) 2^(20)(sin2x-2^(20)sin4x)