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Answer 1 · 2018-02-12T21:07:24

$x = 0, \frac{3}{4} π, π, \frac{7}{4} π$ $x=0, 3/4pi, pi, 7/4pi$

We want to find the solutions of the equation

${sec}^{2} (x) + tan (x) = 1$ $sec^2(x)+tan(x)=1$

Use the trigonmetric identity ${sec}^{2} (x) = {tan}^{2} (x) + 1$ $color(blue)(sec^2(x)=tan^2(x)+1)$

${tan}^{2} (x) + 1 + tan (x) = 1$ $tan^2(x)+1+tan(x)=1$

$\Leftrightarrow {tan}^{2} (x) + tan (x) = 0$ $<=>tan^2(x)+tan(x)=0$

$\Leftrightarrow tan (x) (tan (x) + 1) = 0$ $<=>tan(x)(tan(x)+1)=0$

Therefore

$tan (x) = 0 or tan (x) = - 1$ $tan(x)=0 or tan(x)=-1$

By the inverse tangent with $x \in [0, 2 π)$ $x in [0,2pi)$

$x = 0, \frac{3}{4} π, π, \frac{7}{4} π$ $x=0, 3/4pi, pi, 7/4pi$