Solving Trig Equations, need help.

corsair29bd · December 13th 2006, 04:08 PM

Im stuck on 2 problems:

1. cos2x(2cos+1) (I dont know even where to start)

and

tan3x=1

this one im beter off on

i relize that tan=1 on (pie/4)

diveded by three for the three rotations you get (pie/12)

i know the answers are (pie/12 + pie N) the book says they are (pie/12, 9pie/12, 17pie/12, 5pie/12, 13pie/12, 21pie /12) how did it get these

thanks

**topsquark** · December 13th 2006, 05:02 PM

Originally Posted by corsair29bd

Im stuck on 2 problems:

1. cos2x(2cos+1) (I dont know even where to start)

Is there an equation here to solve?

First off, the symbol $\pi$ is "pi" not "pie."

$cos(2x) = 2cos^2(x) - 1$ so:

$cos(2x) \cdot (2cos(x)+1) = (2cos^2(x) - 1)(2cos(x)+1)$

= $4cos^3(x) + 2cos^2(x) - 2cos(x) - 1$

Now write your equation and we can help you further.

-Dan

**topsquark** · December 13th 2006, 05:11 PM

Originally Posted by corsair29bd

tan3x=1

$tan(3x) = 1$

$3x = tan^{-1}(1) = \frac{\pi}{4}$

Now, this is a reference angle. So actually we have the 2 equations:
$3x = \frac{\pi}{4}$

$3x = \frac{5\pi}{4}$

(We ignore quadrants II and IV since tangent is negative there.)

So $x = \frac{\pi}{12}$ and $x = \frac{5 \pi}{12}$ .

Now, we are dividing by 3 so there should be 3 solutions associated with each x solution we have. (This is because we can add multiples of $2\pi$ to each solution for 3x.)

$x = \frac{\pi}{12}, \frac{\pi}{12} + \frac{2\pi}{3}, \frac{\pi}{12} + \frac{4\pi}{3}$

$x = \frac{5\pi}{12}, \frac{5\pi}{12} + \frac{2\pi}{3}, \frac{5\pi}{12} + \frac{4\pi}{3}$

And this generates the list you gave as a solution.

-Dan

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