Trig Identities and Logs, Solve x within the given interval

**tacoking** · August 14th 2010, 02:52 AM

Note: I wasn't really sure which board to post this on since this involves both trig identities and logs, but I figured that since Trigonometry generally comes after Algebra that I'd post it in this board. If I put this topic in the wrong board, I apologize for my confusion.

Solve for x in the interval [0, 2 $\pi$ ]:

$\log_{2}\cot x - 2\log_{4}\csc2x = \log_{2}\cos x$

Now, so far I've tried turning the $2log_{4}$ into $4\log_{2}$ and then getting stuck with not knowing how to break down $csc 2x$ so I can use log identities (subtracting logs with same bases becomes division of factors, for reference). I've also tried breaking it down to the following, but I hit a dead end because it doesn't really seem to lead anywhere that I can tell:

$\log_{2}csc x = \log_{4}csc4x^2$

The answer is supposed to be $x = \frac{\pi}{3} , \frac{5\pi}{3}$

So, any tips on how to progress, or which way to progress, would be appreciated. Thanks.

**Failure** · August 14th 2010, 04:28 AM

Originally Posted by tacoking

Note: I wasn't really sure which board to post this on since this involves both trig identities and logs, but I figured that since Trigonometry generally comes after Algebra that I'd post it in this board. If I put this topic in the wrong board, I apologize for my confusion.

Solve for x in the interval [0, 2 $\pi$ ]:

$\log_{2}\cot x - 2\log_{4}\csc2x = \log_{2}\cos x$

Now, so far I've tried turning the $2log_{4}$ into $4\log_{2}$

Note that $\log_4(x)=\frac{1}{2}\log_2(x)$

and then getting stuck with not knowing how to break down $csc 2x$

Just write the whole thing with $\sin$ and $\cos$ . Then collect the left side inside a single $\log_2$ , then drop the $\log_2$ on both sides of the equation. Remember that $\sin (2x)=2\sin(x)\cos(x)$ .
After you have done that you get the pretty simple equation
$2\cos(x)=1$

**tacoking** · August 14th 2010, 05:06 AM

I forgot about that trig identity... >.< Thanks much for the help!

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