Question #9e7a0

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Question #9e7a0

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Answer 1 · 2018-02-26T10:57:08

$x = 0, 1.77, 4.51, 2 π$ $x=0,1.77,4.51,2pi$

Explanation:

First, we will use the identity ${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x=sec^2x-1$

${sec}^{2} x - 1 + 4 sec x = 4$ $sec^2x-1+4secx=4$

${sec}^{2} x + 4 sec x - 5 = 0$ $sec^2x+4secx-5=0$

$a = sec x$ $a=secx$

$a^{2} + 4 a - 5 = 0$ $a^2+4a-5=0$

$(a - 1) (a + 5) = 0$ $(a-1)(a+5)=0$

$a = 1 or a = - 5$ $a=1 or a=-5$

$sec x = 1 or sec x = - 5$ $secx=1 or secx=-5$

$cos x = 1 or - \frac{1}{5}$ $cosx=1 or -1/5$

$x = arccos (1) = 0 and 2 π$ $x=arccos(1)=0 and 2pi$
or
$x = arccos (- \frac{1}{5}) \approx {1.77}^{c} or ~ {4.51}^{c}$ $x=arccos(-1/5)~~1.77^c or ~4.51^c$

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Question #9e7a0

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