Question

Solve the following equation

Answer 1

`\bf sec^2(x)-2=0\implies \cfrac{1^2}{cos^2(x)}-2=0\implies \cfrac{1}{cos^2(x)}=2 \\\\\\ \cfrac{1}{2}=cos^2(x)\implies \sqrt{\cfrac{1}{2}}=cos(x) \\\\\\ cos^(-1)\left((1)/(√(2)) \right)=cos^(-1)[cos(x)]\implies cos^(-1)\left((√(2))/(2) \right)=\measuredangle x \\\\\\ \measuredangle x = \begin{cases} (\pi )/(4)\\\\ (7\pi )/(4) \end{cases}`

now, those are the angles from [0, 2π], now, to include "all", using the "n" notation, for all coterminal angles

well, all you do, as before, add 2π or +2π to each

well.... 7π/4 is really -π/4 if you use negative angles

thus
`\bf x=\pm (\pi )/(4)+2\pi n \qquad \textit{ where`

Answer 2

`x= ( \pi )/(4) +2 \pi n, (7 \pi )/(4) +2 \pi n, (3 \pi )/(4) +2 \pi n, (5 \pi )/(4) +2 \pi n`