Question

Is this correct? Solve the equation on the interval of \( 0\geq \theta < 2\pi \)
\( 2cos \theta + 3 = 2 \)
\( 2 cos \theta = 2-3 \)
\( 2 cos \theta = -1 \)
\( \frac{2 cos \theta}{2} = \frac{-1}{2} \)
\( cos \theta = \frac{-1}{2} \)

This was on my midterm and I made an A on it but this one I got wrong but this looks correct? What did I do wrong?

Answer 1

Solve the equation on the interval of \( 0\geq \theta < 2\pi \)
\[ \huge 2\cos \theta + 3 = 2 \]
\[ \huge 2 \cos \theta = 2-3 \]
\[\huge 2 \cos \theta = -1 \]
\[ \huge \frac{2 \cos \theta}{2} = \frac{-1}{2} \]
\[\huge \cos \theta = -\frac{1}{2} \]

Answer 2

wait how many answers did you provide? there are two answers.

Answer 3

One sec it should be \(\huge 0\leq \theta < 2\pi \)

Answer 4

I provided one

Answer 5

but if we work out the equation, like I did, it comes to what I have

Answer 6

You need to solve for \(\sf\theta\)

Answer 7

Ok so he wanted the angle

Answer 8

Yes, and you should get 2 answers for the angle

Answer 9

Yes I get two. I was thinking he just wanted us to do it like that because some of our homework does it like I just did it.

Answer 10

What 2 angles did you get? :)

Answer 11

cos(a) = x on the unit circle with radius 1... picture that thing

Answer 12

I got 120 and 240 degrees. or 2pi/3 and 4pi/3

Answer 13

Correct :)

Answer 14

Yep that is it :-) Just a silly mistake on my part :-( oh well lol

Answer 15

Got the hardest one right because I put the angle lol

Answer 16

If they didn't have the restrictions on the value of \(\sf\theta\)

then the answer would be:

\(\sf\Large \theta=\frac{2\pi}{3}\pm2\pi n\)

\(\sf\Large \theta=\frac{4\pi}{3}\pm2\pi n\)

Answer 17

Funny, I made the same mistake on two but for some reason, on the hardest one I put the angle for that one but not the other two.... . DUFF what was I thinking lol

Answer 18

Anyways... I still have an A in the class and I made an A on the midterm so I am happy :-)

Answer 19

(-: