Question #e6469

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Answer 1 · 2016-06-24T18:47:09

$x=128°$

By multiplying, you obtain:

$(cos^2x-1)/cos^2x=1/cosx$

that is equivalent to:

$(cos^2x-1)/cos^2x=cosx/cos^2x$

and, by fixing $cosx!=0$ ,

$cos^2x-1-cosx=0$

$cos^2x-cosx-1=0$

$cosx=(1+-sqrt(1+4))/2$

$cosx=(1+-sqrt(5))/2$

$cosx=(1+sqrt(5))/2$ is impossible

$cosx=(1-sqrt(5))/2$

$x=128°$