Question

Find the solutions of the equation that are in the interval [0, 2π).

2-2sin(t)=2sqrt(3)cos(t)

Answer 1

Consider squaring both sides and using Pythagorean identity.

Answer 2

Should I factor out the 2 on the left and cancel first?

Answer 3

yeah that could help.

Answer 4

Ok, well, if I square the left I will get sin^2x-2sinx+1, and that isn't very helpful from what I can tell right?

Answer 5

and then on the left have 3cos^2x

Answer 6

sorry for mixing my variable name up

Answer 7

\[
3-3\cos^2(t)=3\sin^2(t)
\]

Answer 8

I'm sorry, I don't understand.
2(1-sint)=2sqrt(3)cost
1-sint=sqrt(3)cost

Now if I square both sides (1-sint)^2 does not equal 1-sin^2t but 1-2sint+sin^2t

Answer 9

I never said it did.

Answer 10

How'd you get 3-3cos^2t=3sin^2t?

Answer 11

It's an equation, you can add it whatever you want to add in if you do it to both sides.

Answer 12

subtract \(3\) from both sides. \(3\cos^2(t)-3 = - (3-3\cos^2(t))\)

Answer 13

???

Answer 14

This looks sketchy without seeing your exact steps

Answer 15

\[\begin{split}
1-2\sin(t)+\sin^2(t)&=3\cos^2(t)\\
-2-2\sin(t)+\sin^2(t)&=3\cos^2(t)-3\\
-2-2\sin(t)+\sin^2(t)&=-(3-3\cos^2(t))\\
-2-2\sin(t)+\sin^2(t)&=-3\sin^2(t)\\
-2-2\sin(t)+4\sin^2(t)&=0&x=\sin(t)\\
4x^2-2x-2&=0
\end{split}\]

Answer 16

Ok! This makes a lot of sense now. I guess I was confusing myself too much by focusing so hard on the left hand side when the right hand side was easier.