Solve for x on $[0, 2 π)$ $[0,2pi)$ : $\frac{x - π}{{cos}^{2} x} < 0$ $(x-pi)/cos^2x < 0$ ?

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Solve for x on $[0, 2 π)$ $[0,2pi)$ : $\frac{x - π}{{cos}^{2} x} < 0$ $(x-pi)/cos^2x < 0$ ?

2 Answers

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Answer 1 · 2018-07-17T13:21:34

$x < π$ $x< pi$ and $x \neq \frac{π}{2}$ $xne pi/2$

Explanation:

It must be ${cos}^{2} (x) \neq 0$ $cos^2(x)ne 0$ since ${cos}^{2} (x)$ $cos^2(x)$ is the denominator of the given inequality and ${cos}^{2} (x) > 0$ $cos^2(x)>0$ in the given interval except $x = \frac{π}{2}$ $x=pi/2$ .
So we get $x < π$ $x < pi$

Answer 2 · 2018-07-17T14:03:29

The solution is $x \in (0, \frac{π}{2}) \cup (\frac{π}{2}, π)$ $x in (0, pi/2)uu(pi/2, pi)$

Explanation:

The inequality is

$\frac{x - π}{{cos}^{2} x} < 0$ $(x-pi)/(cos^2x)<0$

And the interval is $I = [0, 2 π)$ $I=[0,2pi)$

Let

$f (x) = \frac{x - π}{{cos}^{2} x}$ $f(x)=(x-pi)/(cos^2x)$

Build a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $0$ $0$ $a a a a a a$ $color(white)(aaaaaa)$ $\frac{π}{2}$ $pi/2$ $a a a a a$ $color(white)(aaaaa)$ $π$ $pi$ $a a a a$ $color(white)(aaaa)$ $\frac{3}{2} π$ $3/2pi$ $a a a a a$ $color(white)(aaaaa)$ $2 π$ $2pi$

$a a a a$ $color(white)(aaaa)$ $x - π$ $x-pi$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ ${cos}^{2} x$ $cos^2x$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a$ $color(white)(aa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a$ $color(white)(aa)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (x) < 0$ $f(x)<0$ when $x \in (0, \frac{π}{2}) \cup (\frac{π}{2}, π)$ $x in (0, pi/2)uu(pi/2, pi)$

graph{(y-(x-pi)/(cosx)^2)=0 [-16.99, 19.04, -9.44, 8.58]} #

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Solve for x on $[0, 2 π)$ $[0,2pi)$ : $\frac{x - π}{{cos}^{2} x} < 0$ $(x-pi)/cos^2x < 0$ ?

2 Answers

Explanation:

Explanation:

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Solve for x on $[0, 2 π)$ $[0,2pi)$ : $\frac{x - π}{{cos}^{2} x} < 0$ $(x-pi)/cos^2x < 0$ ?