Question

How do you find the exact value of `2tan^2theta+6tantheta=20` in the interval `0<=theta<2pi`?

Answer 1

`theta=63.435^o`, `101.1^o`, `243.435^o` and `281.1^o`.

Explanation:

`2tan^2theta+6tantheta=20`

`hArrtan^2theta+3tantheta-10=0`

or `tan^2theta+5tantheta-2tantheta-10=0`

or `tantheta(tantheta+5)-2(tantheta+5)=0`

or `(tantheta-2)(tantheta+5)=0`

i.e. `tantheta=2` or `-5`.

Hence as observed from tables for tangent or scientific calculators, `theta=63.435^o` and `101.31^o` as also `180^o + 63.435^o=243.435^o` and `180^o + 101.31^o=281.31^o` as tangent ratio has cycle of `pi` or `180^o`.

Answer 2

The answer is `={1.11, 1.77, 4.25, 4.91} rad`

Explanation:

This is a quadratic equation in `tantheta`

You solve this like

`ax^2+bx+c=0`

`2tan^2theta+6tantheta-20=0`

Divide the equation by `2`

`tan^2theta+3tantheta-10=0`

First, we calcuate the discriminant

`Delta=b^2-4ac`

`=3^2-4*1*-10=49`

As `Delta>0`, we have 2 real roots

`tantheta=(-b+-sqrtDelta)/(2a)`

`=(-3+-7)/2`

`tantheta=-5` , and `tantheta=2`

`theta={63.4º,101.3º, 243.4º,281.3º}`

`theta={1.11, 1.77, 4.25, 4.91} rad`