How do you solve tan x/2 + tan2x? help please?

Question

sorry, forgot to put =0 at the end

my bad

so its tan x/2 + tan2x=0

Answer 1

Since you say solve, I'll assume you typed + instead of =. The expression tan(x/2)+tan(2x) can be written purely in terms of x/2, x, or 2x but it's messy and long - I doubt any assignment would be that tedious. Solving tan(x/2)=tan(2x) is actually relatively straight forward:

tan(x/2)=tan(2x)

sin(x/2)/cos(x/2)=sin(2x)/cos(2x)             (since tax(y)=sin(y)/cos(y))

sin(x/2)cos(2x)=sin(2x)cos(x/2)               (multiply up)

sin(2x)cos(x/2)-sin(x/2)cos(2x)=0            (move it all to one side)

sin(2x-x/2)=0              (use identity sin(a-b)=sin(a)cos(b)-cos(a)sin(b) )

sin(3x/2)=0

The locations where the function sin(u) takes the value 0 are where u=k*Pi for some integer k.

so 3x/2 = k*Pi for some integer k

so the solutions are precisely those x=2*k*Pi/3       (k integer)

Answer 2

Unik has explained how to simplify. But is there something on rhs also?

Answer 3

I'm not sure exactly how you would do this particular question but I have found a website with lots of trig identities on which you might find useful:

www.clarku.edu/~djoyce/trig/identities...

Answer 4

isnt that trignometry.

sorry, but im in 7th grade

Answer 5

sub x = 2m

tan4m = sin4m/cos4m

i suspect you have to apply double angle formula

sin2A = 2sinAcosA

cos2A = cos²A - sin²A

sin4m/cos4m = (2.sin2m.cos2m) / [(cos2m.cos2m) - (sin2m.sin2m)] = [2.2sinm.cosm.(cos²m - sin²m)] / [(cos²m - sin²m)² - (4sin²m.cos²m)]

it'll be a long evalutation but i suppose that is how u do. pls correct me if i'm wrong.

thx

Answer 6

You can solve an equation only!!!

For something to be called as eqn, there must be a = sign. Sadly it is missing.

Answer 7

tan x/2=sinx/(1+cosx)=(1 - cosx)/sinx

tan 2x=2tanx/(1 - tan^2x)    

tanx=sinx/cosx

obeying these formula, u can get them simplified.