Inequality

dhiab · #1 $Old$ June 5th, 2009, 02:53 AM

Solve in

:

Deadstar · #2 $Old$ June 5th, 2009, 06:30 AM

Below is a sin graph between $-3\pi$ and $2\pi$ .
The horizontal blue line indicates y=0.5.
The vertical blue lines show where this cuts the x-axis.
What you want is every point on the x-axis that is outside the two blue boxes since you can see that if x takes any value inside either of the blue boxes, the corresponding y-value is greater than 0.5.

Now note that $\sin(x) = \frac{1}{2}$ when $x = \frac{\pi}{6}$ and again when $x = \pi - \frac{\pi}{6}$ . This is the boundary of the box on the positive x-axis and tells us that x cant take any values between $x = \frac{\pi}{6}$ and $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Since sin(x) is $2\pi$ periodic, if we subtract $2\pi$ from the above results we get the values on the negative x-axis that x cant take.

So putting it all together you get that x can take any value in the following range...
$\bigg{[}-3\pi, -\frac{11\pi}{12}\bigg{]}$ , $\bigg{[}-\frac{7\pi}{6}, \frac{\pi}{6}\bigg{]}$ , $\bigg{[}\frac{5\pi}{6}, 2\pi\bigg{]}$

Amer · #3 $Old$ June 5th, 2009, 06:39 AM

first take all intervals when the sinx is negative so you will have

$\pi\leq x\leq 2\pi$
in the negative
$-\pi\leq x\leq 0$
$-3\pi\leq x\leq -2\pi$

now we need all values of x which sinx less than or equal 1/2

$0\leq x\leq\frac{\pi}{6}$
$\frac{5\pi}{6}\leq x\leq \pi$

now in the negative
$\frac{-7\pi}{6}\leq x\leq -\pi$

there is still one interval just I will leave it for you take this graph it will help you ....

I am always late

$inequality-vb.jpg$