Solve for $\theta$ in $2\sec^2θ-4 =0$ Solve for $\theta$ in $2\sec^2θ-4 =0$
I have gotten toward $\sec^2θ=2$ Then $\dfrac{1}{\cos^2θ} = 2$ 
What is the next step to this problem?
 A: Next Steps: $\cos^2\theta=\dfrac{1}{2}$, then $\cos\theta=\pm\dfrac{1}{\sqrt{2}}$.
Going to cosine was a good idea. Alternately, you could use the identitity $1+\tan^2\theta=\sec^2\theta$ to get $2(\tan^2\theta+1)=4$, and then $\tan^2\theta=1$, then $\tan\theta=\pm 1$.
A: $\dfrac{1}{\cos^2\theta}=2 \implies\cos^2 \theta=\dfrac{1}{2}\implies \cos \theta=\pm\dfrac{1}{\sqrt{2}}\implies\theta=\arccos(\pm\dfrac{1}{\sqrt{2}})$ 
A: Method $1:$ 
When we only have $\cos^2\theta$ in the equation, its better to utilize  $\cos2\theta=2\cos^2\theta-1$
As we have $2\cos^2\theta-1=0, \cos2\theta=0$
$\displaystyle\implies 2\theta=(2n+1)\frac\pi2$ where $n$ is any integer
or as we know if $\cos x=\cos \alpha,x=2n\pi\pm \alpha$
$\displaystyle\implies 2\theta=2p\pi\pm\frac\pi2$  where $p$ is any integer

Method $2:$ 
Alternatively, we know  if $\cos^2x=\cos^2A\iff \sin^2x=\sin^2A\iff \tan^2x=\tan^2A$
$\displaystyle x=m\pi\pm A$ where $m$ is any integer
Now, we have $\cos^2\theta=\frac12=\left(\frac1{\sqrt2}\right)^2=\cos^2\frac\pi4$
$\implies x=r\pi\pm \frac\pi4$ where $r$ is any integer
Can you show that all the solutions are equivalent?
A: This is a difference of squares:
$0 = 2 \sec^2 \theta - 4 = 2( \sec \theta + \sqrt2)( \sec \theta - \sqrt2)$
So one of the two factors of the product on the right is $0$. In the first case, $\sec \theta = - \sqrt2$, so $\cos \theta = -\frac{1}{\sqrt2}$, and in the second $\cos \theta = \frac{1}{\sqrt2}$. Each of these possibilities has infinitely many solutions for $\theta$ real, but if you restrict to $\theta$ in the interval $[0,\pi]$ then the unique solutions are given by $\theta = \cos^{-1}(\pm \frac{1}{\sqrt{2}})$.
