Answer:
[tex]x=0,\frac{2\pi}{3},\frac{4\pi}{3}[/tex]Explanation:
Given the trigonometric equation:
[tex]tan^2x+secx=1[/tex]We are required to find the solutions on the interval [0, 2π).
First, recall the trigonometric identity below:
[tex]\tan^2x=\sec^2x-1[/tex]Substitute the identity above for tan²x in the given equation.
[tex]\begin{gathered} tan^{2}x+secx=1 \\ \sec^2x-1+secx=1 \\ \sec^2x-1+\sec x-1=0 \\ \implies\sec^2x+\sec x-2=0 \end{gathered}[/tex]Factorize the resulting quadratic equation:
[tex]\begin{gathered} \sec^2x+2\sec x-\sec x-2=0 \\ \sec x(\sec x+2)-1(\sec x+2)=0 \\ (\sec x-1)(\sec x+2)=0 \end{gathered}[/tex]Thus:
[tex]\begin{gathered} \sec x-1=0,\sec x+2=0 \\ \sec x=1,\sec x=-2 \end{gathered}[/tex]Finally, we solve for x in the interval [0, 2π):
[tex]\begin{gathered} x=\sec^{-1}1=0\in[0,2\pi) \\ x=\sec^{-1}(-2)=\frac{2\pi}{3},\frac{4\pi}{3} \end{gathered}[/tex]The solutions to the equation on the given interval are:
[tex]x=0,\frac{2\pi}{3},\frac{4\pi}{3}[/tex]