Question

@LolWolf Solve 2 sin^2 x = 1 for all values of x such that 0 <= x < 2pi

Answer 1

Note that:
\[
2\sin^2(x)=1\implies\\
\sin(x)=\pm\frac{\sqrt{2}}{2}
\]

Answer 2

Or, simply \(|\sin(x)|=\frac{\sqrt{2}}{2}\), hence, we have that:
\[
x\in\{\frac{\pi}{4}, \frac{3\pi}{4}\}
\]

Answer 3

Sorry, noted it was \(2\pi\), so:
\[
x\in \{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\}
\]

Answer 4

here sinx=\[1/\sqrt{2}\] so x=pi/4 ,3pi/4,5pi/4 and 7pi/4

Answer 5

@LolWolf Set notation would more appropriately be:\[x \in \left\{ \frac{ \pi }{ 4 },\frac{ 3 \pi }{ 4 } \right\} \cup \left\{ \frac{ 5 \pi }{ 4 },\frac{ 7 \pi }{ 4 } \right\}\]

Answer 6

okay that is what I got, but I got there a little differently...

divide both sides by 2 gives you sin^2 x = 1/2
square root to get rid of exponent. write it as arcsin (probably not necessary) and then find 1/ sqrt 2 on unit circle...

Answer 7

Fun stuff!

Answer 8

It's exactly the same thing @genius12 , the union operator makes it so.

As for the case, be careful with \(asin(x)\) as it only returns one value of multiple.
But, yet, simply find \(\frac{\sqrt{2}}{2}\) on the unit circle.