sec^2(x)−1=0

mattgondek4 2021-08-13 Answered
\(\displaystyle{{\sec}^{{2}}{\left({x}\right)}}−{1}={0}\)

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Expert Answer

Theodore Schwartz
Answered 2021-08-14 Author has 16157 answers
\(\displaystyle{{\sec}^{{2}}{\left({x}\right)}}-{1}={0}\)
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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