How do I solve sec^2x + 2 tan^2x =4?

1 Answer
Sep 20, 2015

Considering only the angles from 0 to 2\pi, the solutions are \pi/4 and {5\pi}/4. Both solutions have a periodicity of 2\pi.

Explanation:

Recall the definitions:

  • sec(x)=1/\cos(x)
  • \tan(x)=\sin(x)/\cos(x)

We can thus write sec^2(x) + 2 \tan^2(x) as

1/\cos^2(x) + 2 \sin^2(x)/\cos^2(x)= {1+2\sin^2(x)}/{\cos^2(x)}.

We want this expression to equal 4, so we can multiply for \cos^2(x) and get

{1+2\sin^2(x)}/{\cos^2(x)}=4 \iff 1+2\sin^2(x)=4\cos^2(x).

Subtracting \cos^2(x) from both sides, we have

1-\cos^2(x) +2\sin^2(x)=3\cos^2(x),

and since 1-\cos^2(x) equals \sin^2(x), the expression becomes

3\sin^2(x)=3\cos^2(x).

Dividing the whole expression by the right member, we have

{3\sin^2(x)}/{3\cos^2(x)}=1. Canceling the 3's out, and recalling again that \tan(x)=\sin(x)/\cos(x), we finally find out that the previous equation is the same as

\tan^2(x)=1, which is verified (considering only the angles between 0 and 2\pi) when \tan(x)=1, i.e. when x=\pi/4, and when \tan(x)=-1, i.e. when x={5\pi}/4