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Tan ^2x + 3 tanx + 1
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So what's the question?
Solve the equation in the interval [0, 2pi) using the quadratic equation
let u = tanx u^2 + 3u + 1 = 0 solve for u using the quadratic equation. plug tanx = u back in. let me know if this solves your problem or if i should proceed ^_^
I got -3 +/- square root 5 divided by 2
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D = 3^2 - 4(1)(1) = 9 - 4 = 5
Kira... I do not understand how you get it to equal 0.36468. I keep getting a negative. What am I doing wrong?
For alpha 1 and alpha 2, you need to take tan inverse of the modulus of the value. Because this gives you the basic angle and from there, you manipulate the values. What you found was the argument. |dw:1382930335718:dw|
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