Question

Show how to solve: 2cos x^2 = 1. Answers are pi/4, 3pi/4, 5pi/4, 7pi/4.

Answer 1

divide by 2 as a first step to get \(\cos^2(x)=\frac{1}{2}\) then take the square root, don't forget the \(\pm\)

Answer 2

get \(\cos(x)=\frac{\sqrt{2}}{2}\) or \(\cos(x)=-\frac{\sqrt{2}}{2}\) and these should look familiar

Answer 3

Satellite73 has solved it.

now you find the value of x and repeats the periods then you will get your answer.\[\large x=\cos ^{-1}(\sqrt{1/2})\]\[x=\pi/4\]the cosine function is positive in the 1st and 4th quadrant. To find the second solution subtract the reference angle(2*pi) to find the solution in 4th quadrant.\[\large x=2\pi-\pi/4\]\[\large x=7\pi/4\]

Answer 4

Now again repeat the same procedure with
\[\large x=\cos ^{^{-1}}(-\sqrt{1/2})\]and you will get 3pi/4 and 5pi/4

Good luck.

Answer 5

@Shayaan_Mustafa: But how you derived a general solution for this? ;)

Answer 6

you mean\[\large 7\pi/4+2n \pi\]are you asking for this?

Answer 7

um I got for the +ve part,

\[2n\pi \pm \frac{\pi}{4} \]

and for the negative part,

\[2n\pi \pm \frac{3\pi}{4} \]

Now how to combine this two?

Answer 8

You are getting me right?

Answer 9

where are others two?

Answer 10

Sorry, I don't understand.

Answer 11

and just write the full set of solution will be then
x=....... so on.

Answer 12

Sorry, I am not getting it :(

Answer 13

ok let me dinner then i will again online. till then kindly be patient.

Answer 14

lol, really? :P

Answer 15

@Shayaan_Mustafa: I was talking about the general solution not for this particular problem. And yes I got it now. Simply looking at the unit circle gives the answer.

Answer 16

To be honest there is no need to do any analysis at all.

Answer 17

All odd multiples of \( \frac{\pi}{4}\) gives the solution.

Answer 18

Which answers he OP's query too!!