Question #b3f1c

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Question #b3f1c

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Answer 1 · 2016-11-16T21:07:59

$x \in [\frac{π}{6}, \frac{5 π}{6}]$ $x in [pi/6, (5pi)/6]$

Explanation:

Note first that as $- 1 \leq sin x \leq 1$ $-1 <= sinx <= 1$ for all real $x$ $x$ , we must have $sin x - 2 \leq - 1 < 0$ $sinx-2 <= -1 < 0$ . Then, $(sin x - \frac{1}{2}) (sin x - 2)$ $(sinx-1/2)(sinx-2)$ is the product of $sin x - \frac{1}{2}$ $sinx-1/2$ and a negative number, meaning it is less than or equal to $0$ $0$ if and only if $sin x - \frac{1}{2} \geq 0$ $sinx-1/2 >= 0$ .

Thus, the given problem is equivalent to the problem $sin x - \frac{1}{2} \geq 0$ $sinx-1/2 >= 0$ with the restriction $x \in [0, 2 π]$ $x in [0, 2pi]$

Adding $\frac{1}{2}$ $1/2$ to each side of the inequality, we get

$sin x \geq \frac{1}{2}$ $sinx >= 1/2$

Using the unit circle, we can tell that on our restricted interval, we have $sin x \geq \frac{1}{2}$ $sinx>=1/2$ if and only if $x \in [\frac{π}{6}, \frac{5 π}{6}]$ $x in [pi/6, (5pi)/6]$ . Therefore, this is our answer.

$x \in [\frac{π}{6}, \frac{5 π}{6}]$ $x in [pi/6, (5pi)/6]$

Answer 2 · 2016-11-17T01:18:03

$x \in [\frac{π}{6}, \frac{5 π}{6}]$ $x in [pi/6, (5pi)/6]$

Explanation:

Alternate answer, using the quadratic formula, as requested.

First, find where the left hand side is equal to $0$ $0$ using the quadratic formula:

$(sin x - \frac{1}{2}) (sin x - 2) = 0$ $(sinx-1/2)(sinx-2) = 0$

$\Rightarrow {(sin x)}^{2} - \frac{5}{2} sin x + 1 = 0$ $=> (sinx)^2-5/2sinx + 1 = 0$

$\Rightarrow sin x = \frac{- (- \frac{5}{2}) \pm \sqrt{{(- \frac{5}{2})}^{2} - 4 (1) (1)}}{2 (1)}$ $=> sinx = (-(-5/2)+-sqrt((-5/2)^2-4(1)(1)))/(2(1))$

$= \frac{\frac{5}{2} \pm \sqrt{\frac{9}{4}}}{2}$ $=(5/2+-sqrt(9/4))/2$

$= \frac{\frac{5}{2} \pm \frac{3}{2}}{2}$ $=(5/2+-3/2)/2$

$= \frac{5}{4} \pm \frac{3}{4}$ $=5/4+-3/4$

$\Rightarrow sin x = 2 or sin x = \frac{1}{2}$ $=> sinx = 2 or sinx = 1/2$

(Note that we could have arrived at this conclusion directly from the factored form by setting each factor equal to $0$ $0$ )

As $sin x \neq 2$ $sinx != 2$ for any $x in RR$ , this leaves us with

$sinx = 1/2$

Using knowledge of common angles, a unit circle, or a calculator, we find that $sinx = 1/2$ has the solutions $x in {pi/6, (5pi)/6}$ in the interval $[0, 2pi]$ .

As $(sinx-1/2)(sinx-2)$ is a continuous function, any interval in which it changes sign must include a $0$ . Thus, if we split the interval $[0, 2pi]$ by removing the points ${pi/6, (5pi)/6}$ , we get three intervals:

$[0, pi/6)$ , $(pi/6, (5pi)/6)$ , $((5pi)/6, 2pi]$

and the sign of $(sinx-1/2)(sinx-2)$ is constant on each of these intervals, as none of them contain a $0$ of $(sinx-1/2)(sinx-2)$ .

With that, we can determine the sign of $(sinx-1/2)(sinx-2)$ on each interval by testing a value in each one, and noting that the sign of that value will be the same as the sign of all of the values on that interval.

$0 in [0, pi/6)$ and

$(sin(0)-1/2)(sin(0)-2) = (-1/2)(-2) = 1 > 0$

Thus $(sinx-1/2)(sinx-2) > 0$ on $[0, pi/6)$

$1 in (pi/6, (5pi)/6)$ and

$(sin(1)-1/2)(sin(1)-2) = (1/2)(-1) = -1/2 < 0$

Thus $(sinx-1/2)(sinx-2) < 0$ on $(pi/6, (5pi)/6)$

$pi in ((5pi)/6, 2pi]$ and

$(sin(2pi)-1/2)(sin(2pi)-2) = (-1/2)(-2) = 1 > 0$

Thus $(sinx-1/2)(sinx-2) > 0$ on $((5pi)/6, 2pi]$

Adding in the points at which the function is $0$ , we obtain our final result:

$(sinx-1/2)(sinx-2) <= 0$ if $x in [pi/6, (5pi)/6]$

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