How do I solve sec2x + 2 tan2x =4?

1 Answer
Sep 20, 2015

Considering only the angles from 0 to 2π, the solutions are π4 and 5π4. Both solutions have a periodicity of 2π.

Explanation:

Recall the definitions:

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

We can thus write sec2(x)+2tan2(x) as

1cos2(x)+2sin2(x)cos2(x)=1+2sin2(x)cos2(x).

We want this expression to equal 4, so we can multiply for cos2(x) and get

1+2sin2(x)cos2(x)=41+2sin2(x)=4cos2(x).

Subtracting cos2(x) from both sides, we have

1cos2(x)+2sin2(x)=3cos2(x),

and since 1cos2(x) equals sin2(x), the expression becomes

3sin2(x)=3cos2(x).

Dividing the whole expression by the right member, we have

3sin2(x)3cos2(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

tan2(x)=1, which is verified (considering only the angles between 0 and 2π) when tan(x)=1, i.e. when x=π4, and when tan(x)=1, i.e. when x=5π4