Question #ae8df

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Question #ae8df

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Answer 1 · 2017-03-23T20:48:43

$x = {0, \frac{π}{4}, \frac{3}{4} π, π, \frac{5}{4} π, \frac{7}{4} π, 2 π}$ $x={0,pi/4,3/4pi,pi,5/4pi,7/4pi,2pi}$

Explanation:

${tan}^{8} x - {tan}^{4} x = {tan}^{4} x ({tan}^{4} x - 1) = 0$ $tan^8 x -tan^4 x=tan^4x(tan^4x-1)=0$

so we have

${tan}^{4} x = 0 \to tan x = 0$ $tan^4 x=0->tan x = 0$

and

${tan}^{4} x - 1 = ({tan}^{2} x + 1) ({tan}^{2} x - 1) = ({tan}^{2} x + 1) (tan x - 1) (tan x + 1)$ $tan^4x-1=(tan^2x+1)(tan^2x-1)=(tan^2x+1)(tanx-1)(tanx+1)$

Here ${tan}^{2} x + 1 = 0$ $tan^2x+1=0$ is unfeasible so we will consider only

${(tanx=0),(tanx=1),(tanx=-1):}$

and then

$x = kpi uu x=pi/4+kpiuux=3/4pi+kpi$ for $k=0,1,2,cdots$

or

$x={0,pi/4,3/4pi,pi,5/4pi,7/4pi,2pi}$

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