How do you solve ${sec}^{5} x = 4 sec x$ $sec^5x = 4secx$ from 0 to 2pi?

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How do you solve ${sec}^{5} x = 4 sec x$ $sec^5x = 4secx$ from 0 to 2pi?

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Answer 1 · 2015-08-13T05:05:27

Solve ${sec}^{5} x = 4 sec x$ $sec^5 x = 4sec x$

Ans: $\pm \frac{π}{4}, \pm \frac{3 π}{4})$ $+- pi/4, +- (3pi)/4)$

Explanation:

${sec}^{5} x - 4 sec x = 0$ $sec^5 x - 4sec x = 0$

$sec x ({sec}^{4} x - 4) = sec x ({sec}^{2} x - 2) ({sec}^{2} x + 2) =$ $sec x(sec^4 x - 4) = sec x(sec^2 x - 2)(sec^2 x + 2) =$

$= sec x (sec x - \sqrt{2}) (sec x + \sqrt{2}) ({sec}^{2} x + 2) = 0$ $= sec x(sec x - sqrt2)(sec x + sqrt2)(sec^2 x + 2) = 0$

a. $sec x = \frac{1}{cos x} = 0$ $sec x = 1/cos x = 0$ (undefined)
b. $\frac{1}{cos x} = \sqrt{2}$ $1/cos x = sqrt2$ --> $cos x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $cos x = 1/sqrt2 = sqrt2/2$ --> $x = \pm \frac{π}{4}$ $x = +- pi/4$
c. $sec x = \frac{1}{cos x} = - \sqrt{2}$ $sec x = 1/cos x = -sqrt2$ --> $cos x = - \frac{1}{\sqrt{2}}$ $cos x = -1/sqrt2$ --> $x = \pm \frac{3 π}{4}$ $x = +-(3pi)/4$
d. (sec^2 x + 2) (always positive)

Answers: $\pm \frac{π}{4}, \pm \frac{3 π}{4}$ $+- pi/4, +- (3pi)/4$
Note. $A r c - \frac{π}{4}$ $Arc -pi/4$ is arc $\frac{7 π}{4}$ $(7pi)/4$ , and arc $- \frac{3 π}{4}$ $-(3pi)/4$ is the arc $\frac{5 π}{4}$ $(5pi)/4$

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How do you solve ${sec}^{5} x = 4 sec x$ $sec^5x = 4secx$ from 0 to 2pi?

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How do you solve ${sec}^{5} x = 4 sec x$ $sec^5x = 4secx$ from 0 to 2pi?