How do you find the exact value of #3tan^2theta-4tantheta-1=0# in the interval #0<=theta<2pi#?

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Answer 1 · 2016-12-08T15:20:52

The answer is #={57.1º,237.1º,347.9º,167.9º} #

You solve this like a quadratic equation

#ax^2+bx+c=0#
Where you replce #x# by #tantheta#

We calculate, first the determinant

#Delta=b^2-4ac=(-4)^2-4(3)(-1)#

#=16+12=28#

As, #Delta>0#, we have 2 real roots

#x=(-b+-sqrtDelta)/(2a)#

So,

#tantheta=(4+-sqrt28)/6#

#=(2+-sqrt7)/3#

#theta_1=arctan((2-sqrt7)/3)=347.9#º and #=167.9º#

and #theta_2=arctan((2+sqrt7)/3)=57.1#º and #=237.1º#