1. solving trig equation

ok i'm having trouble with this problem...

i think there must be a very easy way of doing it but i'm just not getting it.....( i ended up in a huge mess)

tan(4x)=1.....sounded really easy to me!

$=(sin(4x))/cos(4x)$

$sin(4x) =2sin(2x)*cos(2x)$

$
sin(4x) = 2*(2sin(x)cos(x))*(cos^2(x)-sin^2(x))
$

now,

$sin4x = (4sinxcosx)*(cos^2x-sin^2x)$

which is

$
4sinx*cos^3x-4sin^3x*cosx
$

and i used the same expansion method to expand cos(4x) and tried to solve but i still don't see the answer which has to be solved algebrically

help is appreaciated...

thnx

2. First off, is there any restrictions for x? Like $0 \le x < 2\pi$ or something? Because that's what I'm going to assume.

You're thinking too hard. Ask your self, the tangent of what angle is 1? It's one of the special angles, and you know that it's $\frac{\pi}{4}$. Tangent is positive in the 3rd quadrant, so another angle would be $\frac{\pi}{4}$. But we're not done.

What we found is that ${\color{red}4}x = \frac{\pi}{4}$ and ${\color{red}4}x = \frac{5\pi}{4}$. But if we go by the restriction $0 \le x < 2\pi$, then it follows that $0 \le {\color{red}4}x < {\color{red}8}\pi$. This means that there are a lot more angles whose tangent is 1. Since the period of tangent is $\pi$, we can just take $4x = \frac{\pi}{4}$ and keep adding $\pi$ until we find all angles in $0 \le {\color{red}4}x < {\color{red}8}\pi$. The complete list of angles is therefore
$4x = \frac{\pi}{4},\;\frac{5\pi}{4},\;\frac{9\pi}{4},\; \frac{13\pi}{4},\;\frac{17\pi}{4},\;\frac{21\pi}{4 },\;\frac{25\pi}{4},\;\frac{29\pi}{4}$.

Now take each angle and divide by 4 to get our x values:
$x = \frac{\pi}{16},\;\frac{5\pi}{16},\;\frac{9\pi}{16} ,\;\frac{13\pi}{16},\;\frac{17\pi}{16},\;\frac{21\ pi}{16},\;\frac{25\pi}{16},\;\frac{29\pi}{16}$.

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