Question #ef759

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Answer 1 · 2017-11-23T07:28:06

Please see below.

We have an identity that says:

#tan^2x+1=sec^2x#.

Let's plug this in for #sec^2x#:

#3(tan^2x+1)=4tanx+7#

#3tan^2x+3-4tanx-7=0#

#3tan^2x-4tanx-4=0#

You can factor this or use the quadratic formula to solve for #tanx#. if you factor it you will get:

#(3tanx+2)(tanx-2)=0#

From #3tanx+2=0# we get #tanx=-2/3# which gives us:

#x=arctan(-2/3)# or #x=-33.7 Degrees#

From #tanx-2=0# we get #tanx=2# which gives us:

#x=arctan(2)# or #x=63.43 Degrees#