# sec^2(x)−1=0

$$\displaystyle{{\sec}^{{2}}{\left({x}\right)}}−{1}={0}$$

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Theodore Schwartz
$$\displaystyle{{\sec}^{{2}}{\left({x}\right)}}-{1}={0}$$
Let: $$\displaystyle{\sec{{\left({x}\right)}}}={i}$$
$$\displaystyle{u}^{{2}}-{1}={0}$$
$$\displaystyle{u}^{{2}}-{1}+{1}={0}+{1}$$
$$\displaystyle{u}^{{2}}={1}$$
$$\displaystyle{u}=\sqrt{{1}},{u}=-\sqrt{{1}}$$
u=1, u=-1
Substitute back $$\displaystyle{u}={\sec{{\left({x}\right)}}}$$
$$\displaystyle{\sec{{\left({x}\right)}}}={1},{\sec{{\left({x}\right)}}}=-{1}$$
$$\displaystyle{x}={2}п{n},{x}=п+{2}п{n}$$
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Answer is given below (on video)