# sec2x-1=0 [0, 2pi) solve for x

$$\displaystyle{\sec{{2}}}{x}-{1}={0}{\left[{0},{2}\pi\right)}$$
solve for x

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Demi-Leigh Barrera
$$\displaystyle{\left(\frac{{1}}{{\cos{{2}}}}{x}\right)}-{1}={0}$$
$$\displaystyle{\left(\frac{{1}}{{\cos{{2}}}}{x}\right)}-{\left({\cos{{2}}}\frac{{x}}{{\cos{{2}}}}{x}\right)}={0}$$
$$\displaystyle\frac{{{1}-{\cos{{2}}}{x}}}{{\cos{{2}}}}{x}={0}$$
$$\displaystyle{1}-{\cos{{2}}}{x}={0}$$
$$\displaystyle{1}-{\left({\left({\cos{{x}}}\right)}^{{2}}-{\left({\sin{{x}}}\right)}^{{2}}\right)}={0}$$
$$\displaystyle{1}-{\left({\cos{{x}}}\right)}^{{2}}+{\left({\sin{{x}}}\right)}^{{2}}={0}$$
$$\displaystyle{\left({\sin{{x}}}\right)}^{{2}}+{\left({\sin{{x}}}\right)}^{{2}}={0}$$
$$\displaystyle{2}{\left({\sin{{x}}}\right)}^{{2}}={0}$$
$$\displaystyle{\left({\sin{{x}}}\right)}^{{2}}={0}$$
$$\displaystyle{\sin{{x}}}={0}$$
$$\displaystyle{x}={0}$$ or $$\displaystyle{x}=\pi$$