Question

Solve the equation in the interval [0, 2π)
cos2x=√2-cos2x

Answer 1

move -cos2x to left side, what u get

Answer 2

ur question, is cos2x=√2 - cos2x or cos2x=√(2-cos2x)

Answer 3

@RadEn it's cos2x=√2 - cos2x. and for the first question it's 2cos2x

Answer 4

ok, it can be
2cos2x = √2 or
cos2x = 1/2 * √2 , right ?

Answer 5

@RadEn yes. that's what i did, but i stopped there

Answer 6

just remember, what's angles when cos of that angle has the value 1/2 * √2 ?

Answer 7

it's the special degrees

Answer 8

π/4

Answer 9

any else?

Answer 10

pi/4 just in the first quadrant, but in the 4th quadrant is .... degreess

Answer 11

and 7π/4

Answer 12

yes, correct

Answer 13

so, the equation can be :
case I :
cos2x = 1/2 * √2
cos2x = cospi/4
cos2x = cos(2pi*n +- pi/4)
so, 2x = 2pi*n +- pi/4 or
x = (2pi*n +- pi/4)/2 = pi*n +- pi/8 (this is just general solution)
do same idea, for
case II :
cos2x = 1/2 * √2
cos2x = cos7pi/4
continue it....

Answer 14

the answers will be {π/8, 9π/8, 7π/8, 15π/8}

Answer 15

yes, for the 2nd case happened duplicates for solution from the 1st case
u are right

Answer 16

i understand it now. thank you so much @RadEn :)

Answer 17

very welcome :)