How do you find $sec x = \sqrt{2}$ $sec x= sqrt2$ ?

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How do you find $sec x = \sqrt{2}$ $sec x= sqrt2$ ?

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Answer 1 · 2018-06-05T21:23:05

Recall the definition of $sec x$ $secx$ in terms of ratios of sides of a right-angled triangle: $sec x = \frac{h y p}{a d j}$ $secx=(hyp)/(adj)$ - hypotenuse divided by adjacent sides.

Consider a triangle with other angles both $45^{o}$ $45^o$ . The two sides adjoining the right-angle are of equal length, call it $a$ $a$ . Pythagoras' Theorem tells us that the hypotenuse is of length $\sqrt{a^{2} + a^{2}} = a \sqrt{2}$ $sqrt(a^2+a^2)=asqrt2$ . So ${sec 45}^{o} = \frac{a \sqrt{2}}{a} = \sqrt{2}$ $sec45^o=(asqrt2)/a=sqrt2$ .

If this doesn't seem intuitive due to the use of the less usual $sec$ $sec$ trig function, consider the definition of $sec x$ $secx$ as $sec x = \frac{1}{cos x}$ $secx=1/cosx$ . $sec x = \sqrt{2} \Rightarrow cos x = \frac{1}{\sqrt{2}}$ $secx=sqrt2rArrcosx=1/sqrt(2)$ , one of the first cosine values that one learns: ${cos 45}^{o} = \frac{1}{\sqrt{2}}$ $cos45^o=1/sqrt2$ .

Answer 2 · 2018-06-05T22:06:16

$x = \frac{π}{4}, \frac{7 \cdot π}{4}$ $x=pi/4,(7*pi)/4$

Explanation:

This can be more simply interpreted if you change $sec x$ $secx$ to $\frac{1}{cos x}$ $1/cosx$ .

Then by manipulating the equation you can turn:
$\frac{1}{cos x} = \sqrt{2}$ $1/cosx=sqrt(2)$

Into:
$cos x = \frac{1}{\sqrt{2}}$ $cosx=1/sqrt(2)$

Since you cannot have radicals in the denominator, you multiply $\frac{1}{\sqrt{2}}$ $1/sqrt(2)$ by $\frac{\sqrt{2}}{\sqrt{2}}$ $sqrt(2)/sqrt(2)$

Giving you: $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$

From here, you go back to the unit circle. Where does the x value (due to cosine) equal $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ ?

When $θ$ $theta$ (in this case x) is equal to $\frac{π}{4}$ $pi/4$

Since $sec x = \sqrt{2}$ $secx=sqrt(2)$ is positive and you are looking for an x value, you must remember that quadrants 1 and 4 have positive x values.

knowing that your answer is $θ$ $theta$ (again, in this case x) $= \frac{π}{4}$ $=pi/4$ , assuming that this is being taken from $[0, 2 π]$ $[0,2pi]$ , makes your two answers $\frac{π}{4}$ $pi/4$ and $\frac{7 \cdot π}{4}$ $(7*pi)/4$

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How do you find $sec x = \sqrt{2}$ $sec x= sqrt2$ ?

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