Mathematics OpenStudy (jana):

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

8 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.

8 years ago OpenStudy (jana):

thank you !

8 years ago OpenStudy (shadowfiend):

No problem! Hope it was enough help :)

8 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 $$\frac{\sqrt{2}}{2}$$.

8 years ago OpenStudy (jana):

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

8 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}

8 years ago OpenStudy (jana):

ohhhh ok haha thanks

8 years ago OpenStudy (shadowfiend):

No problem, glad to help :)

8 years ago
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