Question

Find all solutions of the equation 2sin2x−cosx=1 in the interval [0,2π), what is x?

Find all solutions of the equation 2sin2x−cosx=1 in the interval [0,2π), what is x?

@Mathematics

Answer 1

hey its actually 2sin^(2)x- cosx=1

Answer 2

That changes things. sin^2 x = 1 - cos^2 x
So you can rewrite your equation as a quadratic equation in cos x.

Answer 3

sin^2 x - cos^2x + 1?

Answer 4

\[ 2\sin^2x−\cos x = 2(1 - \cos^2 x) + \cos x = -2\cos^2 x + \cos x + 2 \]

Answer 5

Hence if \[ 2\sin^2 x−\cos x = 1 \]
then
\[ -2\cos^2 x + \cos x + 1 = 0 \]

Answer 6

Now let \( u = \cos x \), then you solve the quadratic equation
\[ 2u^2 - u - 1 = 0 \]
for u and then find the x such that cos x equals those solutions.

Answer 7

whoa, do you think you can demonstrate that. im not sure i know how to solve for u and x there

Answer 8

You can solve \( 2u^2 - u - 1 = 0 \). Use the quadratic formula. Tell me what solutions you get.

Answer 9

(1,-.5)

Answer 10

x1= 1
x2= -.5

Answer 11

Correct: \( u = 1, -1/2 \).

Now, for what x is
\[ \cos x = 1 \]or
\[ \cos x = -1/2 \]
where \( x \ \in \ [0,2\pi) \).

Answer 12

is that saying x does not equal to 0,2pi

Answer 13

Where x is in that interval. That's what you have as part of your original question; x is in that interval.

Answer 14

so if i had to enter an answer for x, i would say what?

Answer 15

"Find all solutions of the equation 2sin^2x−cosx=1 in the interval [0,2π)"

Answer 16

For what x is cos x = 1?
and for cos x = -1/2?

Answer 17

this is exactly what my homework says
Find all solutions of the equation 2sin2x−cosx=1 in the interval [0,2π).
The answer is x= .
Note: If there is more than one solution, enter them separated by commas. If needed enter π as pi.

Answer 18

i dont know what they want me enter for x...should i say x is in that interval

Answer 19

So far we have shown that cos x = 1 or cos x = -1/2.

Now the question is for what x is that true.

For example, for what values of x is it the case that cos x = 1?

Answer 20

how do i find that?

Answer 21

What is sin 0 ?

Answer 22

0 right

Answer 23

yes. What is sin(pi/2)?

Answer 24

1

Answer 25

Yes. For values of x is sin x = 0?

Answer 26

yes

Answer 27

Find all solutions of the equation sec^(2)x−2=0.
The answer is A+Bkπ where k is any integer and 0<A<π/2,
Find A and B

can you help me find just B also pls James

Answer 28

You might want to fine-tune your solution, as the equation you solve should be
2u^2 + u -1=0 , u =cos(x). I think you flipped a sign somewheres

Answer 29

phi its is not accpeting the u in the equation

Answer 30

Thanks Phi. Yes, if
\[ 2\sin^2 x - \cos x = 1 \]
then
\[ 2 (1 - \cos^2 x) - \cos x = 1 \]
\[ -2\cos^2 x - \cos x + 1 = 0 \]
\[ 2u^2 + u - 1 = 0 \ \ \ \ \ ---- (*)\]
where \( u = \cos x \)
Now from the equation (*), \( u = 1/2 \ or -1 \).
Now for what x is cos x = 1/2 or cos x = -1?

Answer 31

If x is in the interval [0,2pi),
cos x = -1 <=> x = pi

cos x = 1/2 <=> pi/3, 5pi/3

Therefore the final answer is
x=π/3,π,5π/3.

Answer 32

@Jinnie, you should sketch out cos(x) between 0 and 2pi, and then mark all the places it takes on the above values. This way you won't miss any of the solutions

Answer 33

oh wow james i had a near exact answer yesterday but i had 3pi/4 somewhere in there

Answer 34

Find all solutions of the equation sec^(2)x−2=0. The answer is A+Bkπ where k is any integer and 0<A<π/2, Find A and B


can you help me find just B
thats all i need

Answer 35

\[ \sec^2 x - 2 = 0 \iff \cos x = \pm 1/\sqrt{2}\]
Now solve these seperately,
cos x = 1/sqrt(2) <=> x = ...what?
cos x = -1/sqrt(2) <=> x = ...what?

Answer 36

this is what i ended up with when i tried to do this
sec^(2) x−2=0
1/ cos^2 x = 2
2cos^2 x = 1
cos^2 x = 1/2
cos x = (+/-)1/√2
x = π/4 x= 3π/4
x=5π/4 x =7π/4

x =π/4 + 2kπ
x= 3π/4 + 2kπ
x=5π/4 + 2kπ
x = 7π/4 + 2kπ

Answer 37

The only thing I don't understand is how you can express all those solutions with A between 0 and pi/2??

Answer 38

\[ x = \pi/4 + k\pi/2 \]