Question

Solve the equation for exact solutions over the interval [0, 2π).sec^2x-2=tan^2x

Answer 1

Step-by-step explanation

We want to find the solution to the equation:

`\sec ^2x-2=\tan ^2x`

First, let us rewrite the trigonometric terms:

`(1)/(\cos^2x)-2=(\sin ^2x)/(\cos ^2x)`

Multiply both sides of the equation by cos²x:

`\begin{gathered} (\cos^2x)/(\cos^2x)-2\cos ^2x=\sin ^2x \\ 1-2\cos ^2x=\sin ^2x \end{gathered}`

We have that:

`\cos ^2x+\sin ^2x=1`

Substitute that for 1 in the equation:

`\cos ^2x+\sin ^2x-2\cos ^2x=\sin ^2x`

Simplify the equation above:

`\begin{gathered} \cos ^2x-2\cos ^2x=\sin ^2x-\sin ^2x \\ \Rightarrow-\cos ^2x=0 \end{gathered}`