How do you solve #3tan^3x = tan x# in the interval 0 to 2pi?

Question

43574 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-13T04:38:17

so
so #x = 30 ,210,150,330,0,180, 360#

Divide both sides by #tan x#

# 3 tan^2 x = 1#

# tan^2 x = 1/3#

# sqrt(tan^2 x) = sqrt(1/3#

#tan x =+- 1/sqrt 3#

Now we have the interval # 0 - 2pi = 0^@ - 360^@#

Lets look at our unit circle;

#tan x =+- 1/sqrt 3#
Lets see which values work for this
so #x = 30 ,210,150,330#

#Tanx# can also = 0

so #x = 30 ,210,150,330,0,180, 360#

Answer 2 · 2016-03-13T04:42:38

We have that

#3tan^3x = tan x=>tanx*(3*tan^2x-1)=0=> tanx=0 or tanx=+-sqrt3/3#

From #tanx=0=>x=0,x=pi,x=2pi#

From #tanx=sqrt3/3=>x=pi/6, x=7/6pi#

From #tanx=-sqrt3/3=>x=5/6*pi , x=11*pi/6#

Hence the solutions for #x# in #[0,2pi]# are

#0,pi,2pi,pi/6,5/6*pi,7/6*pi,11/6*pi#