Question #75006

We need

#sectheta=1/costheta#

#cos^2theta+sin^2theta=1#

Dividing by #cos^2theta#

#1+sin^2theta/cos^2theta=1/cos^2theta#

#1+tan^2theta=sec^2theta#

The equation

#2sec^2theta+3tantheta=1#

becomes

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

#2+2tan^2theta+3tantheta-1=0#

#2tan^2theta+3tantheta+1=0#

We solve this like the quadratic equation

#ax^2+bx+c=0#

The discriminant is

#Delta=b^2-4ac=9-(4*2*1)=9-8=1#

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

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

#tantheta=(-3+-sqrt1)/(2*2)#

#tantheta=-1#

#theta=3/4pi# or #theta=135º#

and

#tantheta=-0.5#

#theta=2.68# or #theta=153.4º#

Use trig identity:
#sec^2 t= (1 + tan^2 t)#
In this case:
#(2 + 2tan^2 t) + 3tan t - 1 = 0#
#2tan^2 t + 3tan t + 1 = 0#
Solve this quadratic equation for tan t.
Since a - b + c = 0, use shortcut.
The 2 real roots are : tan t = - 1 and tan t = -c/a = - 1/2
Use calculator and unit circle:
a. tan t = -1 --> #t = (3pi)/4# and #t = pi + (3pi)/4 = (7pi)/4#
b. #tan t = - 1/2# --> #t = -26^@57#, and
#t = -26.57 + 180 = 153^@43#
Answers for #(0, pi)#:
#(3pi)/4 (or 135^@); 153^@43#