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Find exact solutions for the interval [0,2pi)?

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  • Find exact solutions for the interval [0,2pi)?


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Answer #1 | 14/01 2014 10:08
1-cos^2 x + 2cos x= -2 cos^2(x) - 2cos(x) -3 = 0 http://www.wolframalpha.com/input/?i=sin%5E2+x+%2B+2cosx%3D-2
Positive: 75 %
Answer #2 | 14/01 2014 10:42
step by step solution: http://symbolab.com/solver/trigonometric-equation-calculator/%20sin%5E%7B2%7D(x)%20%2B%202cos(x)%3D-2%2C%200%5Cle%20x%3C%202%5Cpi hope this helps
Positive: 75 %
Answer #3 | 14/01 2014 10:15
remember your trig identies sin^2x+cos^2x=1 so subtracting cos^2x from both sides sin^2x=1-cos^2x so 1-cos^2x+2cosx=-2 adding 2 to both sides we get 1-cos^2x+2cosx+2=0 multiply everything by -1 cos^2x-2cosx-3=0 this is a quadratic we can factorize this ( cosx-3)(cosx+1)=0 cosx-3=0 then cosx=3 this isn't possible because cosx cannot be greater than 1 cosx+1=0 then cosx=-1 so x= arccos(-1) x=pi
Positive: 66.666666666667 %

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