tanØ = sinØ/cosØ = 0
in sinØ/cosØ = 0 , that is only possible if sinØ = 0
and you should know that the basic sine curve has zeros at 0 , π and 2π
since the period of sinØ = π,
a new answer can be obtained by simply adding/subtracting multiples of π (or nπ) to any anwer
So why not start with 0 + nπ = nπ
Solve the trig equation exactly over the indicated interval.
tanθ = 0, all real numbers
I know the answer is πn, but I don't understand how they got this. I get that tanθ = 0 on π and 2π on the unit circle, so did they just put n on the end of π because of this?
2 answers
Use trigonometric identities to solve tan(2θ)+tan(θ)=0 exactly for 0≤θ≤π. If there is more than one answer, enter your answers as a comma separated list
0 answers