Question

how do you solve 2sin^(2)2x=1

Answer 1

So, first off, you can do the usual stuff to isolate the \(\sin 2x\): divide by 2 and then take the square root.

\[\sin 2x = \sqrt{\frac{1}{2}}\]

Now, you should know from the unit circle which number's sin is \(\frac{1]{2}\) -- and you can set \(2x =\) that.

Answer 2

thank you !

Answer 3

No problem! Hope it was enough help :)

Answer 4

Er, sorry, I messed up -- you need to find out which number's sin is \(\sqrt{\frac{1}{2}}\) -- which is the same as \(\frac{\sqrt{2}}{2}\).

Answer 5

how is \[\sqrt{(1/2)}\] the same as \[\sqrt{(2)/2}\]?

Answer 6

Not the same thing :)

\[\sqrt{\frac{1}{2}}\]

is the same as

\[\frac{\sqrt{2}}{2}\]

Note how the square root is over all of 1/2, but only the 2 in 2/2. This is because:

\[\begin{align}
\sqrt{\frac{1}{2}} &= \frac{\sqrt{1}}{\sqrt{2}}\\
&= \frac{1}{\sqrt{2}}\\
&= \frac{1\sqrt{2}}{\sqrt{2}\sqrt{2}}\\
&= \frac{\sqrt{2}}{\left(\sqrt{2}\right)^2}\\
&= \frac{\sqrt{2}}{2}
\end{align}\]

Answer 7

ohhhh ok haha thanks

Answer 8

No problem, glad to help :)