Question

how to solve cos^3x csc^3x tan^3x=csc^2x-cot^2x ?

Answer 1

`x=0,pi,2pi,pi/4,5pi/4`

Explanation:

`cos3xtimes1/(sin3x)timestan3x`=`1/sin(2x)-cos(2x)/sin(2x)`

`1=(1-cos2x)/sin(2x)`

`sin(2x)=(1-cos(2x))`

`2sinxcosx=(cosx)^2+(sinx)^2-(cosx)^2+(sinx)^2`

`2sinxcosx=2(sinx)^2`

`(sinx)^2-sinxcosx=0`

`sinx(sinx-cosx)=0`

`sinx=0` or `sinx-cosx=0`

`sinx=0` or `tanx=1`

`x=0,pi,2pi,pi/4,5pi/4`

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• I'm giving the answers between 0 and `2pi`

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Answer 2

Verified below

Explanation:

Using:
`cscx=1/sinx`
`cotx= 1/tanx`
`1+cot^2x= csc^2x`

Start:
`cos^3x csc^3x tan^3x=csc^2x-cot^2x`

`cos^3x/sin^3x*tan^3x=csc^2x-cot^2x`

`cot^3 tan^3x=csc^2x-cot^2x`

`1/(cancel(tan^3x))* cancel(tan^3x)=csc^2x-cot^2x`

`1= csc^2x-cot^2x`

`1+csc^2x-csc^2x= csc^2x-cot^2x`

`1+csc^2x-(1+cot^2x)= csc^2x-cot^2x`

`cancel(1)+csc^2xcancel(-1)-cot^2x=csc^2x-cot^2x`

Graph of `csc^2x-cot^2x`
graph{(cscx)^2-(cotx)^2 [-9.96, 10.04, -5.04, 4.96]} = csc^2x-cot^2x#

Graph of `cos^3x csc^3x tan^3x`
graph{(cosx)^3(cscx)^3(tanx)^3 [-9.96, 10.04, -5.04, 4.96]}