Question

Solve cos2x=1/√(2)
Interval= 0≤x<2π
Please show steps!

Answer 1

don't just give the answer @nj1202 show him how to do it.

Answer 2

So first, we should state:
\[\cos(\frac{ \pi }{ 4}) =\frac{ \sqrt{2} }{ 2 }\]

So since our function multiplies the angle by 2, then ours would be true as follows:
\[\cos2(\frac{ \pi }{ 8 }) = \frac{ \sqrt{2} }{ 2 }\]

So now, we can find where pi/8 appears in a circle AND where cos(x) is positive!

The phrase for all quadrants: All Students Take Calculus
So cos(x) is positive in the 1st Q and the 4th Q.

So pi/8 in the 4th Q is at 7pi/8.

The answers are: \[\frac{ \pi }{ 8 } , \frac{ 7\pi }{ 8 }\]

Answer 3

Got it, thanks!

Answer 4

how about u @nj1202 , if x=9pi/8
right to, isn't it?

Answer 5

RadEn, you would be correct, however that is the same exact thing at x = pi/8 anyway and we're only looking from 0 to 2pi.

Answer 6

the finally, that's right equal to pi/8...
but i think 9pi/8 satisfy in interval 0<=x<=2pi

Answer 7

Oh, 9pi/8 disregarding we multiply by 2. I understand what you mean now. So yes, 9pi/8 is in the interval, but it's in the 3rd Q where cos(x) is negative... sqrt(2)/2 is positive, so it wouldn't work.

Answer 8

but the problem is cos(2x), not cosx... lol