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**Lloyd****Guest**

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

**Flowers4Carlos****Member**- Registered: 2005-08-25
- Posts: 106

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

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**austin81****Member**- Registered: 2005-03-21
- Posts: 39

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.

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