How can this be solved?

#3tan^2(x) + 13tan(x) - 10#

2 Answers
maganbhai P.
Aug 1, 2018

Note:To complete equation we have to take,
#3tan^2x+13tanx-10=0#

#x={kpi+arctan(2/3) ,kin ZZ}uu{kpi-arctan(5) ,k in ZZ}#

Explanation:

Here ,

#3tan^2x+13tanx-10=0#

#=>3tan^2x-2tanx+15tanx-10=0#

#=>tanx(3tanx-2)+5(3tanx-2)=0#

#=>(3tanx-2)(tanx+5)=0#

#=>3tanx-2=0 or tanx+5=0#

#=>tanx=2/3 or tanx=-5#

#(i)tanx=2/3=>x=kpi+arctan(2/3) ,kin ZZ#

#(ii)tanx=-5=>x=kpi+arctan(-5),kinZZ#

#i.e. x=kpi-arctan(5) ,k in ZZ#

Note:To complete equation we have to take,

#3tan^2x+13tanx-10=0#

Lucy
Aug 1, 2018

#tanx=2/3#

#x=0.588+npi# where #n# is an integer

#tanx=-5#

#x=1.77 +npi# where #n# is an integer

Explanation:

#3tan^2x+13tanx-10#

Think of #tanx# as #y#

So, #3tan^2x+13tanx-10# becomes #3y^2+13y-10#

Then, using the quadratic formula,

#y=(-13+-sqrt(13^2-4(3)(-10)))/(2times3)#

#y=(-13+-sqrt289)/6#

#y=(-13+-17)/6#

#y=(-13+17)/6# and #y=(-13-17)/6#

#y=2/3# or #y=-5#

Since #y=tanx#, then

#tanx=2/3# or #tanx=-5#

#tanx=2/3#

#x=0.588+npi# where #n# is an integer

#tanx=-5#

#x=1.77 +npi# where #n# is an integer

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