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solve the following equations for θ, in the interval 0<θ≤ 360°
a) (sin θ - 1)(5cos θ +3) = 0
&
b) tan θ= tan θ(2+3 sin θ)
Last edited by serena (2009-03-12 22:52:23)
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a) (sin θ - 1)(5cos θ +3) = 0
= 5sinθ cosθ +3sinθ - 5cosθ - 3 = 0
= 5cosθ (sinθ -1) = 3(1-sinθ)
= 5cosθ = -3.
cosθ = -3/5
b) tan θ= tan θ(2+3 sin θ)
Cancelling tanθ on the LHSand RHS,
1 = 2 + 3 sinθ
-1 = 3 sinθ
sinθ = -1/3
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
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thnx a lot
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