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jana
 5 years ago
how do you solve 2sin^(2)2x=1
jana
 5 years ago
how do you solve 2sin^(2)2x=1

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shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0So, 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.

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0No problem! Hope it was enough help :)

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Er, sorry, I messed up  you need to find out which number's sin is \(\sqrt{\frac{1}{2}}\)  which is the same as \(\frac{\sqrt{2}}{2}\).

Jana
 5 years ago
Best ResponseYou've already chosen the best response.0how is \[\sqrt{(1/2)}\] the same as \[\sqrt{(2)/2}\]?

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0Not 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: \[\begin{align} \sqrt{\frac{1}{2}} &= \frac{\sqrt{1}}{\sqrt{2}}\\ &= \frac{1}{\sqrt{2}}\\ &= \frac{1\sqrt{2}}{\sqrt{2}\sqrt{2}}\\ &= \frac{\sqrt{2}}{\left(\sqrt{2}\right)^2}\\ &= \frac{\sqrt{2}}{2} \end{align}\]

shadowfiend
 5 years ago
Best ResponseYou've already chosen the best response.0No problem, glad to help :)
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