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`