Solving for x in a Trigonometric Equation: Where Did I Go Wrong?

[majinknight]

Ok now this question i got wrong too but i got 2 of the answers correct. ok question is solve for x the following equation.
2sin squared x= 1-cosx

Ok this is what i did.
2sin Squared x- sinx= 0
Let u= sinx
2uSquared - u=0
u(2u-1)=0
u=0, u=one half.
sinx=0, sinx=1/2
Now x= o and 2pie. And for the other part it get x=pie/6 and 5pie/6. Now i checked my answers with the sheet as it says answers and it says the answers are 0, 2pie, 2pie/3, 4pie/3. Now 2pie/3 is square root of 3/2 and 4pie/3 is -square root3/2. What did i do wrong?

 

[Tron3k]

I think the problem in your solution is that you assumed that:

1 - \cos x = \sin x

I would do something like this...

2 \sin^2 x = 1 - \cos x
2 (1 - \cos^2 x) = 1 - \cos x
2 - 2 \cos^2 x - 1 + \cos x = 0
-2 \cos^2 x + \cos x + 1 = 0
2 \cos^2 x - \cos x - 1 = 0

Then solve the quadratic equation.

 

[Tron3k]

Even better, you can factor it:

2 \cos^2 x - \cos x - 1 = 0
2 \cos^2 x - 2 \cos x + \cos x - 1 = 0
2 \cos x (\cos x - 1) + (\cos x - 1) = 0
(2 \cos x + 1)(\cos x - 1) = 0

 

[majinknight]

Ya ok i see what i did wrong ,i like to use sin and find myself always using it and i should have went to cousine, i figured out to factor it once you helped me get to that step so thank you so much. I only have 7 more practise questions on this sheet and only 2 have stumped me so i am very proud. Thanks so much and if i need more help again i will ask. Thanks