how do you solve cos^4x-1=cot^2 csc^2x-csc^2x?
I can't for the life of me figure out how to solve this problem and I know its probably really easy. If I could just get a step by step it'd be very helpful.
Update:its a trig proof so im looking to set them equal to each other
More Questions From This User See All
Answers & Comments
Verified answer
cos^4(x) - 1 = cot^2(x) csc^2(x) - csc^2(x)
=> cos^4(x) - 1 = csc^2(x) [ cot^2(x) - 1 ]
=> cos^4(x) - 1 = (cot^2(x) + 1)[ cot^2(x) - 1 ]
=> cos^4(x) - 1 = cot^4(x) - 1
= > cos^4(x) = cot^4(x)
=> cos^4(x) = cos^4(x) / sin^4(x)
=> sin^4(x) cos^4(x) = cos^4(x)
=> cos^4(x) [ sin^4(x) - 1 ] = 0
case 1: cos^4(x) = 0
cos x = 0
x = (2n + 1)Π/2
case 2: sin^4(x) - 1 = 0
sin^4(x) = 1
sin x = 1, -1
x = (2n + 1)Π/2
n is any integer including zero
RHS = cot^2 csc^2x-csc^2x = csc^2 x(cot^2 x- 1) = (cot^2 x+ 1)(cot^2 x-1) = cot^4 x -1
so we get cos^4x-1 = cot^4 x - 1
or cos^4 x = cot ^4 x = cos^4 x/ sin ^4 x
or cos^4 x(1- 1/ sin ^4 x) = 0
cos^4 x= 0 or (1- 1/sin ^4 x) = 0
cos x = 0 => x = (n+1/2) pi
or sin x = 1 or - 1 so x= (n+1/2) pi
so x= (n+1/2) pi