How do you solve $tan(x)(tanx+1)=0$ ?

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Answer 1 · 2018-03-29T05:44:10

$x=0,(3pi)/4, pi,(7pi)/4,,2pi, (11pi)/4, 3pi,(15pi)/4,....$

$tanx(tanx+1)=0$

$tanx=0$

$x=0,pi,2pi,3pi,....$

$tanx+1=0$

$tanx=-1$

$x=pi-pi/4, 2pi-pi/4, 3pi-pi/4,4pi-pi/4,...$

$x=(3pi)/4, (7pi)/4, (11pi)/4, (15pi)/4,....$

Combining and arranging in the increasing order

$x=0,(3pi)/4, pi,(7pi)/4,,2pi, (11pi)/4, 3pi,(15pi)/4,....$

How do you solve tan(x)(tanx+1)=0tan(x)(tanx+1)=0?