1. ## Solve for θ

Solve for all + values of θ. If 0<θ<270

3 - sec^2 2θ - tan^2 2θ = 0

2. Originally Posted by reiward
Solve for all + values of θ. If 0<θ<270

3 - sec^2 2θ - tan^2 2θ = 0
Remember that $\displaystyle \sec^2{x} = \tan^2{x} + 1$.

So

$\displaystyle 3 - (\tan^2{2\theta} + 1) - \tan^2{2\theta} = 0$

$\displaystyle 2 - 2\tan^2{2\theta} = 0$

$\displaystyle 2 = 2\tan^2{2\theta}$

$\displaystyle 1 = \tan^2{2\theta}$

$\displaystyle \pm 1 = \tan{2\theta}$

$\displaystyle 2\theta = \left\{45^{\circ}, 135^{\circ}, 225^{\circ}, 315^{\circ}\right\} + 360^{\circ}n, n \in \mathbf{Z}$

$\displaystyle \theta = \left\{22.5^{\circ}, 67.5^{\circ}, 112.5^{\circ}, 157.5^{\circ}, 202.5^{\circ}, 247.5^{\circ}\right\}$ in the domain $\displaystyle 0^{\circ} \leq \theta \leq 270^{\circ}$.

3. oh sorry, my equation was wrong

it should be

3 - sec^2 2θ - tan 2θ = 0

4. Originally Posted by reiward
oh sorry, my equation was wrong

it should be

3 - sec^2 2θ - tan 2θ = 0

$\displaystyle 3 - (\tan^2{2\theta} + 1) - \tan{2\theta} = 0$
$\displaystyle \tan^2 2\theta+\tan 2\theta-2=0$
$\displaystyle (\tan 2\theta-1)(\tan 2\theta+2)=0$