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

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Answer 1 · 2018-04-03T02:29:54

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

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

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

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

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

$2sinxcosx=2(sinx)^2$

$(sinx)^2-sinxcosx=0$

$sinx(sinx-cosx)=0$

$sinx=0$ or $sinx-cosx=0$

$sinx=0$ or $tanx=1$

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

Answer 2 · 2018-04-03T03:37:09

Verified below

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]}