How do you solve tan 3x(tanx-1)=0tan3x(tanx−1)=0 in the interval 0 to 2pi?

1 Answer
Shwetank Mauria
Feb 13, 2016

Solution set is {0, pi/4, pi/3, (2pi)/3, pi, (5pi)/4, (4pi)/3, (5pi)/3, 2pi}{0,π4,π3,2π3,π,5π4,4π3,5π3,2π}

Explanation:

As tan3x(tanx−1)=0tan3x(tanx−1)=0, this means

Either tan3x=0tan3x=0 i.e.

3x=0 or pi or 2pi or 3pi or 4pi or 5pi or 6pi3x=0orπor2πor3πor4πor5πor6π

i.e. x=0 or pi/3 or (2pi)/3 or pi or (4pi)/3 or (5pi)/3 or 2pix=0orπ3or2π3orπor4π3or5π3or2π

Alternatively (tanx−1)=0(tanx−1)=0 i.e. tanx=1tanx=1

i.e. x=pi/4 or (5pi)/4x=π4or5π4

Hence Solution set is {0, pi/4, pi/3, (2pi)/3, pi, (5pi)/4, (4pi)/3, (5pi)/3, 2pi}{0,π4,π3,2π3,π,5π4,4π3,5π3,2π}

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