SOLUTION: Can you help me to solve Trig equations algebraically? 3tan^3(t)-3tan^2(t)-tan(t)+1=0 What I have so far, if correct;I changed my tan(t) to (x) for less confusion [3tan^3(t)-3ta

Question 672588: Can you help me to solve Trig equations algebraically?
3tan^3(t)-3tan^2(t)-tan(t)+1=0
What I have so far, if correct;I changed my tan(t) to (x) for less confusion
[3tan^3(t)-3tan^2(t)] -tan(t)+1=0
(3x^3-3X^2)-1x+1 = 0
3x(x^2-x)-1(x+1)= 0 pull out the common factor and then stuck!
I should have two answers such as 3x-1 which would actually be 3tan(t)-1, then I would take 3tan(t)=1 which gives me 3tan equals {1+h(2pi)}/3 and {pi-1m(2pi)}/3
I just can't figure this problem further, trig is tough for me =, this may be a the best website ever!
thank you for your time
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Can you help me to solve Trig equations algebraically?
3tan^3(t)-3tan^2(t)-tan(t)+1=0
What I have so far, if correct;I changed my tan(t) to (x) for less confusion
[3tan^3(t)-3tan^2(t)] -tan(t)+1=0
(3x^3-3X^2)-1x+1 = 0
3x(x^2-x)-1(x+1)= 0 pull out the common factor and then stuck!
---------------
Make it
3x^2(x - 1)-1(x+1)= 0
= 0

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