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Topic review (newest first)

austin81
2005-11-06 01:52:16

I'll continue from
where the inv of tan:
sin,cos and tan are +ve at 1st quandrant(angle=@°),sin +ve at 2nd(angle=180°-@),tan+ve at 3rd(angle=180°+@°) and cos+ve at 4th(angle=360°-@°).
Invtan(-1/2)=26.6° and Intan(-3°)=71.6°
Therefore required angles are
(180°-26.6)°, (360-26.6)°,(180-71.6)°and(360-71.6)°
Sorry I don't haveagood programthat illustrate Maths symbols.

Flowers4Carlos
2005-11-05 14:18:10

hi yaz lloyd!

i can't remember how to do these exactly, but i think it should go something like this:

2tan²θ + 7tanθ = -3

2tan²θ + 7tanθ + 3 = 0

2(tan²θ + 7/2tanθ + 3/2) = 0

2(tanθ + 1/2)(tanθ + 3) = 0

tanθ = -1/2           tanθ = -3
θ = tan-¹(-1/2)      θ = tan-¹(-3)

hmm... i guess u need a calculator to find θ

Lloyd
2005-11-05 11:20:37

hello, i need a little help again on trig. please explain this to me: thanks, lloyd

Solve for θ the quadratic 2tan^2 θ + 7tanθ = -3 where 0<θ<360

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