How do i solve this trig equation? tan(3Θ)+1=0 where 0

Question

How do i solve this trig equation? tan(3Θ)+1=0 where 0<=Θ<2π

anonymous · Answer

can you divide the 3 out of the tanΘ?

mathmale · Answer

There are several ways, but I think one of the easier to follow is to replace 3 theta temporarily with x and solve the resulting equation:  tan x + 1

anonymous · Answer

would you have to change the restriction to $$0\le3\theta <6\Pi$$

mathmale · Answer

that is, tan x + 1 + 0.    for what angle or angles, between 0 and 2pi, is tan x = -1?

anonymous · Answer

the reference angle for 1 is pi/4, and tan is negative is quadrants 2 and 4 so it is 3pi/4 and 7pi/8?

mathmale · Answer

would you have to change the restriction to 
0≤3θ<6Π ?     Let's solve the problem; after which I'd bet you could answer your own question.

mathmale · Answer

tan 3Pi/4 is indeed equal to -1.  I'm not convinced that 7Pi/8 is a solution; have you checked it in tan x + 1 = 0?

anonymous · Answer

ohh i meant 7pi/4

anonymous · Answer

so then you set that equal to tan(3Θ) and divide by 3

mathmale · Answer

You probably remember that the period of the tangent function is pi (not 2pi as with the sine and cosine).  

Almost.   Set 3pi/4 equal to 3 theta and solve for theta.

Then do the same for 7pi/8.  checking your own answer (value(s) of theta) will tell you whether you're right that 7pi/4 is a solution of tan x + 1 = 0.

anonymous · Answer

ok, thanks!

mathmale · Answer

My pleasure.  Shall i assume we're done?

anonymous · Answer

yes!

mathmale · Answer

Good for you.  Hope to work with you again.