OpenStudy (jana):

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

7 years ago
OpenStudy (shadowfiend):

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.

7 years ago
OpenStudy (jana):

thank you !

7 years ago
OpenStudy (shadowfiend):

No problem! Hope it was enough help :)

7 years ago
OpenStudy (shadowfiend):

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

7 years ago
OpenStudy (jana):

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

7 years ago
OpenStudy (shadowfiend):

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

7 years ago
OpenStudy (jana):

ohhhh ok haha thanks

7 years ago
OpenStudy (shadowfiend):

No problem, glad to help :)

7 years ago
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