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

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

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Answer 1 · 2018-04-03T02:29:54

$x = 0, π, 2 π, \frac{π}{4}, 5 \frac{π}{4}$ $x=0,pi,2pi,pi/4,5pi/4$

Explanation:

$cos 3 x \times \frac{1}{sin 3 x} \times tan 3 x$ $cos3xtimes1/(sin3x)timestan3x$ = $\frac{1}{sin (2 x)} - \frac{cos (2 x)}{sin (2 x)}$ $1/sin(2x)-cos(2x)/sin(2x)$

$1 = \frac{1 - cos 2 x}{sin (2 x)}$ $1=(1-cos2x)/sin(2x)$

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

$2 sin x cos x = {(cos x)}^{2} + {(sin x)}^{2} - {(cos x)}^{2} + {(sin x)}^{2}$ $2sinxcosx=(cosx)^2+(sinx)^2-(cosx)^2+(sinx)^2$

$2 sin x cos x = 2 {(sin x)}^{2}$ $2sinxcosx=2(sinx)^2$

${(sin x)}^{2} - sin x cos x = 0$ $(sinx)^2-sinxcosx=0$

$sin x (sin x - cos x) = 0$ $sinx(sinx-cosx)=0$

$sin x = 0$ $sinx=0$ or $sin x - cos x = 0$ $sinx-cosx=0$

$sin x = 0$ $sinx=0$ or $tan x = 1$ $tanx=1$

$x = 0, π, 2 π, \frac{π}{4}, 5 \frac{π}{4}$ $x=0,pi,2pi,pi/4,5pi/4$

I'm giving the answers between 0 and $2 π$ $2pi$

Answer 2 · 2018-04-03T03:37:09

Verified below

Explanation:

Using:
$csc x = \frac{1}{sin x}$ $cscx=1/sinx$
$cot x = \frac{1}{tan x}$ $cotx= 1/tanx$
$1 + {cot}^{2} x = {csc}^{2} x$ $1+cot^2x= csc^2x$

Start:
${cos}^{3} x {csc}^{3} x {tan}^{3} x = {csc}^{2} x - {cot}^{2} x$ $cos^3x csc^3x tan^3x=csc^2x-cot^2x$

$\frac{{cos}^{3} x}{{sin}^{3} x} \cdot {tan}^{3} x = {csc}^{2} x - {cot}^{2} x$ $cos^3x/sin^3x*tan^3x=csc^2x-cot^2x$

${cot}^{3} {tan}^{3} x = {csc}^{2} x - {cot}^{2} x$ $cot^3 tan^3x=csc^2x-cot^2x$

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

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

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

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

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

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]}

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

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