How do you use the quadratic formula to solve #2tantheta(tantheta+1)=3# for #0<=theta<360#?

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Answer 1 · 2016-12-23T18:06:28

#theta={39.45^@, 118.75^@, 219.45^@, 298.75^@}#

#2tantheta(tantheta+1)=3#

#hArr2tan^2theta+2tantheta=3#

or #2tan^2theta+2tantheta-3=0#

Using quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#, we get

#tantheta=(-2+-sqrt(2^2-4xx2xx(-3)))/(2xx2)#

= #(-2+-sqrt(4+24))/4#

= #-2/4+-sqrt28/4#

= #-1/2+-sqrt7/2#

= #-0.5+-1.3229#

= #0.8229# or #-1.8229#

Hence #theta=39.45^@# or #theta=118.75^@#

Other solutions are #theta=219.45^@# or #theta=298.75^@#