Solve for [tex]\( x \)[/tex]:

[tex]\[ 3x = 6x - 2 \][/tex]



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[tex]$
\sin \left(\frac{3 \pi}{4}\right)=
$[/tex]
[tex]$\qquad$[/tex]
A. [tex]$\frac{1}{2}$[/tex]
B. [tex]$\frac{\sqrt{2}}{2}$[/tex]
C. [tex]$-\frac{\sqrt{2}}{2}$[/tex]
D. [tex]$\frac{\sqrt{3}}{2}$[/tex]
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Response:
[tex]\[ \sin \left(\frac{3 \pi}{4}\right) = \][/tex]

A. [tex]\(\frac{1}{2}\)[/tex]
B. [tex]\(\frac{\sqrt{2}}{2}\)[/tex]
C. [tex]\(-\frac{\sqrt{2}}{2}\)[/tex]
D. [tex]\(\frac{\sqrt{3}}{2}\)[/tex]



Answer :

To solve for [tex]\(\sin \left(\frac{3 \pi}{4}\right)\)[/tex], we'll need to determine the sine of the given angle, [tex]\(\frac{3 \pi}{4}\)[/tex].

Firstly, it helps to understand where [tex]\(\frac{3\pi}{4}\)[/tex] radians lies on the unit circle. This angle is in the second quadrant, as it is between [tex]\(\frac{\pi}{2}\)[/tex] and [tex]\(\pi\)[/tex]. The reference angle in this case is [tex]\(\pi - \frac{3\pi}{4} = \frac{\pi}{4}\)[/tex].

In the second quadrant, the sine function is positive. Therefore, the value of [tex]\(\sin\left( \frac{3\pi}{4} \right) \)[/tex] will be the same as the sine of the reference angle [tex]\(\frac{\pi}{4}\)[/tex], but positive since sine is positive in the second quadrant.

We know that the sine of [tex]\(\frac{\pi}{4}\)[/tex] is [tex]\(\frac{\sqrt{2}}{2}\)[/tex].

Therefore, [tex]\(\sin \left(\frac{3 \pi}{4}\right)\)[/tex] is also [tex]\(\frac{\sqrt{2}}{2}\)[/tex].

Comparing this with the given options:

A. [tex]\(\frac{1}{2}\)[/tex]
B. [tex]\(\frac{\sqrt{2}}{2}\)[/tex]
C. [tex]\(-\frac{\sqrt{2}}{2}\)[/tex]
D. [tex]\(\frac{\sqrt{3}}{2}\)[/tex]

We see that [tex]\(\sin \left(\frac{3 \pi}{4}\right) = \frac{\sqrt{2}}{2}\)[/tex].

Hence, the correct answer is:
[tex]\[ \boxed{\frac{\sqrt{2}}{2}} \][/tex]