How do I go about solving cos(2x) = 0? Finding all solutions 0<= x <= 2pi?

Question

I start of like this
cos(2x) = cos^2x - sin^2x = 0
but I'm way off according to the book......My professor isn't very good at explaining this to me...I try to get it but there are so many rules to memorize. If someone could please go into detail on how to solve the problem I would GREATLY appreciate it.

Answer 1

Let's ignore the 2x part for now. Cosine of what is 0?

Let's do a substitution just to make things easier to follow for now.

Let u = 2x

Solving u for x: x = u/2

Now your original equation becomes:

cos(u) = 0

The domain changes as well.

0 <= x <= 2pi

0 <= u/2 <= 2pi

Multiply each term by 2 to get rid of the fraction.

0 <= u <= 4pi

Now let's figure out where cos(u) = 0 in the given domain.

Looking at my key points, cos(u) = 0 when u = pi/2 or when u = 3pi/2. We need to go around to 4pi, so we also consider u=pi/2+2pi = 5pi/2 and u=3pi/2 + 2pi = 7pi/2.

u element {pi/2, 3pi/2, 5pi/2, 7pi/2}

2x element {pi/2, 3pi/2, 5pi/2, 7pi/2}

Divide everything by 2.

x element of {pi/4, 3pi/4, 5pi/4, 7pi/4}



I hope this helps.

Answer 2

cos u = 0 when u = pi/2 + n* pi, where n is an integer



if we let u = 2x, the zeros will be 2x = pi/2 + n * pi



or, x = pi/4 + n * pi/2 = pi/4 + n * 2pi/4



solve this for values of n that make x in the interval [0 , 2pi]

n = 0, 1,2,3 (4 solutions)

the period of cos 2x is pi, so there will be 2 complete cycles between 0 and 2pi, thus 4 solutions (2 solutions per cycle)



x = pi/4, 3pi/4 , 5pi/4 , 7pi/4

Answer 3

Solve Cos 2x

Answer 4

cos(2x)=0

this will imply cos^-1(0)=2x

i.e cos inverse of 0 equals 2x

now think of values when cosine of any angle is zero

cosine of pi/2 is 0 and cosine of 3pi/2 is 0 on the interval of 0 to 2pi

no other values of cosine will be zero!!

now cos^-1 (0)is replaced by the values

that means pi/2 equals 2x which implies x equals pi/4

and same as the other one it only has two values!!!

the idea may be different than your teacher but we can use any idea to get the answer

Answer 5

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Use the inverse cosine: 2x = arccos(-0.32) Find the primary solutions: 2x = 1.897, 2x = 4.387, 2x = 8.180, 2x = 10.67 (I obtain the second value by using a cosine graph and its properties) Divide both sides by 2 to get x by itself. x = 0.948, x = 2.19, x = 4.09, x = 5.33 (3 s.f.) General solution to find cosax = b where a and b are constants is therefore: x = [arccos(b)]/a, x = [2π - arccos(b)]/a, x = [arccos(b) + 2π]/a, x= [4π - arccos(b)] etc. I'm not too sure but that's what I make of it! :D

Answer 6

Try solving cos(x) = 0. I'm thinking pi/2, and 3pi/2.



So:

2x = pi/2 and 2x = 3pi/2

x = pi/4 and x = 3pi/4

Answer 7

I guess you're taking pre-cal right now.

an easy way to do,

you can use double angle formula

cos2x=2cos^2x-1

when cos2x=0

2cos^2x-1=0

cos^2x=1/2

cosx=+or-square root of (2)/2

taking ArcCos +square root of (2)/2

which give you,

x=pi/4

taking ArcCos -square root of (2)/2

x=3pi/4

which are the answers on that interval.

Answer 8

cos(2x) = 0 where 2x = π/2, 3π/2, 5π/2, 7π/2, and, in general, (2k+1) * π / 2 where n is any integer.



When 2x = π/2, dividing by 2 to find x we find that:



x = π/4



So each of the answers that I gave for 2x must be divided by 2.



π/4, 3π/4, 5π/4, and 7π/4.



The next answer would have been 9π/4 and that is outside of the chosen interval. Likewise -π/4 is outside of that interval so those are the only solutions.



x = π/4, 3π/4, 5π/4, and 7π/4.....<<<<<...Answer

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