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# maths question watch

1. if you have a trigonemtrical quadratic equation such as

(tanx)^2 +tanx - 1 how do you solve it.
2. Complete the square:

(tanx)^2 +tanx - 1 = 0
(tan x + 1/2 )^2 - 5/4 = 0
(tan x + 1/2 )^2 = 5/4
tan x + 1/2 = ±√(5/4)
tan x = ±√(5/4) - 1/2
x = arctan [ ±√(5/4) - 1/2]
3. this is for pure 3. we dont use arc tan in p3 so there must be another method.
4. (Original post by Tennis)
this is for pure 3. we dont use arc tan in p3 so there must be another method.
x^2 + x -1

solve that to get

x = 0.618 or x = -1.618

so tan x = 0.618 or tanx = -1.618

do it from there..
5. (Original post by Tennis)
this is for pure 3. we dont use arc tan in p3 so there must be another method.
Err why don't you use it? It gives the solutions to the problem, unless you fancy trial and error.
6. (Original post by mik1a)
Err why don't you use it? It gives the solutions to the problem, unless you fancy trial and error.
is arctan...tan-1??
7. what is 1/tan, no thats cot
8. (Original post by lgs98jonee)
is arctan...tan-1??
well is it??
9. Arctan is inverse tan, cot is 1/tan

so if tan x = y
y = arctan x

so yes, acrtan = tan^-1

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