How would I solve the following equation in the interval [0,2#pi#): #3 tan theta + sec^2 theta = 2# ?

Question

Solve the following equation in the interval [0,2#pi#): #3 tan theta + sec^2 theta = 2#

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Answer 1 · 2018-02-05T23:31:20

See below.

Identity:

#color(red)(sec^2(x)=1+tan^2(x))#

#3tan(theta)+1+tan^2(theta)-2=0#

#tan^2(theta)+3tan(theta)-1=0#

Let: #u=tan(theta)#

#:.#

#u^2+3u-1=0#

#(-3+-sqrt(9+4))/2#

#u=(-3+sqrt(13))/(2)# , #u=(-3+sqrt(13))/(2)#

#tan(theta)=u=(-3+sqrt(13))/(2)# ,#theta=arctan((-3+sqrt(13))/(2))#

#=color(blue)(0.29402,pi+0.29402)#

#tan(theta)=u=(-3-sqrt(13))/(2)# ,#theta=arctan((-3-sqrt(13))/(2))#

#=color(blue)(pi-1.2768, 2pi-1.2768)#