Math  /  Trigonometry

Questioncos((1/2)(x))=1/2\cos ((1 / 2)(x))=1 / 2 where xx is a value between 0 and 2pi radians

Studdy Solution

STEP 1

What is this asking? Find the angles xx between 00 and 2π2\pi radians where the cosine of half the angle, x2\frac{x}{2}, equals 12\frac{1}{2}. Watch out! We're dealing with *half* the angle, xx, not xx itself!
Also, make sure your answer is in radians, not degrees.

STEP 2

1. Find solutions for half the angle
2. Find solutions for the full angle
3. Check the range

STEP 3

We're given cos(x2)=12\cos\left(\frac{x}{2}\right) = \frac{1}{2}.
So, we need to figure out what x2\frac{x}{2} could be!

STEP 4

Think about the unit circle!
Where is the cosine equal to 12\frac{1}{2}?
That happens at π3\frac{\pi}{3} and also at 5π3\frac{5\pi}{3}.

STEP 5

Remember, we can keep going around the unit circle!
So, the general solutions for x2\frac{x}{2} are π3+2πk\frac{\pi}{3} + 2\pi k and 5π3+2πk\frac{5\pi}{3} + 2\pi k, where kk is any integer.
This represents all the possible angles.

STEP 6

Now, let's find xx.
If x2=π3+2πk\frac{x}{2} = \frac{\pi}{3} + 2\pi k, then multiplying both sides by 22 gives us x=2π3+4πkx = \frac{2\pi}{3} + 4\pi k.
Similarly, if x2=5π3+2πk\frac{x}{2} = \frac{5\pi}{3} + 2\pi k, then x=10π3+4πkx = \frac{10\pi}{3} + 4\pi k.

STEP 7

Let's try different values of kk to see which ones give us solutions between 00 and 2π2\pi.

STEP 8

If k=0k = 0, we get x=2π3x = \frac{2\pi}{3} and x=10π3x = \frac{10\pi}{3}. 2π3\frac{2\pi}{3} is definitely between 00 and 2π2\pi.
But 10π3\frac{10\pi}{3} is bigger than 2π2\pi, so it's outside our desired range.

STEP 9

What if kk is negative?
If k=1k = -1, we'd get negative values for xx, which are also outside our range.

STEP 10

So, the only solution between 00 and 2π2\pi is x=2π3x = \frac{2\pi}{3}.
It's within the range, and we've considered all possibilities!

STEP 11

The solution to cos(x2)=12\cos\left(\frac{x}{2}\right) = \frac{1}{2} for 0x2π0 \le x \le 2\pi is x=2π3x = \frac{2\pi}{3}.

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