Question

Help me find the solutions in the given domain 0 less than and equal to x<2pi for the equation: cos(2x)=1/sqrt2?

Answer 1

Let me make sure I'm getting this right. The domain is:\[0 \le x \le 2\pi\]
Right? Keep in mind that \[\frac{ 1 }{ \sqrt{2} } = \frac{ \sqrt{2} }{ 2 }\]

Answer 2

Should be able to solve for x, keeping in mind (man I say that a lot) it's not going to be one of the standard angles on the unit circle since its 2*x

Answer 3

\[0 \le x < 2 \pi\]

Answer 4

There should be two solutions to the problem, both when x is positive on the unit circle.

Answer 5

So, it would be 7pi/4 and pi/4 then, right?

Answer 6

Thats where cos(x) is equal to those values, but you need 2x.

Answer 7

How would you solve for 2x then?

Answer 8

Close. But its 2*x. so, at pi/4 in this equation, you're really doing 2*(pi/4) = pi/2

Answer 9

2x = pi/4, 7pi/4
Divide by 2.

Answer 10

First note:\[\bf \frac{ 1 }{ \sqrt{2} }=\frac{ \sqrt{2} }{ 2 }\]Now in the interval \(\bf 0 \le x < 2 \pi\), cos(2x) hits \(\bf 1/\sqrt{2}\) twice as many times as cos(x). We know that for cosine to be \(\bf 1/\sqrt{2}\), the angle inside the brackets must be in the form \(\bf \pi/2 + 2 \pi k\). Hence:\[\bf 2x=\frac{ \pi }{ 4 }+ 2 \pi k \implies x = \frac{ \pi }{ 8 }+ \pi k\]

Answer 11

x = inversecos(1/sprt(2)/2, inversecos(1/sprt(2)/2 + PI/2
x = PI/4, PI/4 + PI/2

Answer 12

\[2x=\frac{ \pi }{ 4 }\]
Solve for x like any other equation.

Answer 13

Now plug in integer values of 'k' to find all values of 'x' between 0 and 2pi. @pokemonmaster96

Answer 14

genius solution is much prettier and more general than mine. If you can remember that, then you could apply that for any equation, not just this one.

Answer 15

I forgot to add, since the angle can actually be pi/8 or 7pi/8 then:\[\bf x=\frac{ \pi }{ 8 }+ \pi k \ \ or \ \ x=\frac{ 7 \pi }{ 8 } + \pi k\]

Answer 16

And the only values of k that satisfy the inequality are k = 0 and k = 1. PLug them in for k in to both and you'll get 4 solutions. @pokemonmaster96

Answer 17

Nope. This is cos(2x) not cos(x).

Answer 18

I got 4 solutions, pi/8, 9pi/8, 7pi/4, and 3pi/4.
Right? @genius12

Answer 19

@pokemonmaster96 The first 2 are right. The other 2 are wrong. Look again at what I wrote 2-3 posts above.

Answer 20

pi/4 and 5pi/4?
If not, then, I am completely lost.

Answer 21

@genius12 ^^^

Answer 22

@pokemonmaster96 The other two solutions would be 7pi/8 and 15pi/8.

Answer 23

2x = inversecos(1/sprt(2), inversecos(1/sprt(2) + PI/2
x = inversecos(1/sprt(2)/2, 2 inversecos(1/sprt(2) + PI/2

Answer 24

Since, it could be two angles, right? The, pi/8 and 7pi/8.
@genius12