Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you solve the following equation #tan 3x = 1# in the interval [0, 2pi]?

1 Answer

Shwetank Mauria

Jun 27, 2016

#x={pi/12,(5pi)/12,(3pi)/4,(13pi)/12,(17pi)/12,(7pi)/4}#

Explanation:

#tan3x=1=tan(pi/4)#

Hence #3x=pi/4# but as function #tanx# has a periodicity of #pi#,

#3x# can take values #{pi/4,(5pi)/4,(9pi)/4,(13pi)/4,(17pi)/4,(21pi)/4,(25pi)/4#

and #x# can take values #{pi/12,(5pi)/12,(9pi)/12,(13pi)/12,(17pi)/12,(21pi)/12}# (we have removed last value as it is beyond #2p#).

or #x={pi/12,(5pi)/12,(3pi)/4,(13pi)/12,(17pi)/12,(7pi)/4}#

Related questions

See all questions in Solving Trigonometric Equations

Impact of this question

19992 views around the world

You can reuse this answer
Creative Commons License