# Thread: Solving Trig Equations, need help.

1. ## Solving Trig Equations, need help.

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

2. 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 $\displaystyle \pi$ is "pi" not "pie."

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

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

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

-Dan

3. Originally Posted by corsair29bd
tan3x=1
$\displaystyle tan(3x) = 1$

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

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

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

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

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

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 $\displaystyle 2\pi$ to each solution for 3x.)

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

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

And this generates the list you gave as a solution.

-Dan