How do I solve $sec (3 x) - \sqrt{2} = 0$ $sec(3x)-sqrt2 = 0$ algebraically?

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Answer 1 · 2018-04-28T15:46:15

$x = \frac{π}{12} + \frac{2}{3} π n$ $x= pi/12+2/3pin$
$x = \frac{7 π}{2} + \frac{2}{3} π n$ $x= (7pi)/2+2/3pin$
n is an element of all integers

Let $u$ $u$ be $3 x$ $3x$ :
$sec (u) - \sqrt{2} = 0$ $sec(u)-sqrt2=0$

$sec u = \sqrt{2}$ $secu= sqrt2$

Apply reciprocal identity:
$\frac{1}{cos u} = \sqrt{2}$ $1/cosu= sqrt2$

$cos u = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $cosu= 1/sqrt2= sqrt2/2$

$u = \frac{π}{4} + 2 π n$ $u= pi/4 +2pin$
$u = \frac{7 π}{4} + 2 π n$ $u= (7pi)/4+2pin$

Now replace $u$ $u$ with $3 x$ $3x$ and solve for x:

$3 x = \frac{π}{4} + 2 π n$ $3x=pi/4 +2pin$
$3 x = \frac{7 π}{4} + 2 π n$ $3x=(7pi)/4+2pin$

$x = \frac{π}{12} + \frac{2}{3} π n$ $x= pi/12+2/3pin$
$x = \frac{7 π}{2} + \frac{2}{3} π n$ $x= (7pi)/2+2/3pin$

Here's a graph:
graph{sec(3x)-sqrt2 [-10, 10, -5, 5]}

How do I solve sec(3x)−√2=0sec(3x)-sqrt2 = 0 algebraically?