Question

2cos2x=-sinx
solve for x over the interval [-pi,+pi]

x= 0.6349radians
x=2.507radians
x= 1.003 radians
x= 2.138 radians

Answer 1

2cos2x=-sinx
2cos 2x + sinx = 0
2( 1 - 2sin^2 x) + sinx = 0
4sin^2 x - sinx - 2 = 0
sinx = (1 ± √1 - 4(4)(-2)) /8
= (1 ± √33)/8
= appr .84307.... or appr -.59307...

If sinx = .84307
x is in quads I or II, and x = appr 1.002967 or appr. 2.1386

if sinx = -.59307..
then x is in quads III or IV
x = 3.776459.. which is beyond the given domain
or
x = 5.6483, beyond the given domain
how about 5.6483 - 2π= -.634867 ?? , that fits the domain. , you did not have that.
how about 3.776459 - 2π = -2.5067 ? YUP, that works also.

so x = appr. 1.002967, 2.1386, -.634867, and -2.5067

as confirmed by:
http://www.wolframalpha.com/input/?i=y+%3D+2cos+(2x)+%2B+sinx

showing 2 positve answers and 2 negative answers in
[-π,+π]

(I also checked these 4 answers with a calculator. You have shown 4 answers that are all positive. The graph contradicts that)

Answer 2

2cos2x=-sinx
2cos 2x + sinx = 0
2( 1 - 2sin^2 x) + sinx = 0
4sin^2 x - sinx - 2 = 0

it should put
4sin^2 x - sinx + 2 = 0

Answer 3

so if c=2
then the values have to be x= 0.6349radians
x=2.507radians
x= 1.003 radians
x= 2.138 radians

Answer 4

If calc mean calculus then:

2 cos ( 2x ) = - sinx

2 cos ( 2x ) + sinx = 0


f(x) = sin x + 2 cos ( 2x )

f´(x) = cos x - 4 cos ( 2x )


x(n+1) = xn - [ sin xn + 2 cos ( 2xn ) ] / [ cos xn - 4 cos ( 2 xn ) ]

This is very, very difficult for calculate.

If you in wolframalpha. c o m type:

solve 2 cos( 2x ) = - sin ( x ) for x = - pi to pi

The solutions are:

-0.634867

-2.50673

1.00297

2.13863

In your list only x= 1.003 radians and x= 2.138 radians are correct solutions.

Answer 5

Nope, my equation is correct, I should not have skipped that one step:

2cos2x=-sinx
2cos2x + sinx = 0
2(1 - 2sin^2 x) + sinx = 0
2 - 4sin^2 x + sinx = 0
times -1
-2 + 4sin^2 x - sinx = 0
4sin^2 x - sinx - 2 = 0

checking one of your answers in the original equation:
let x = 2.5067
Left Side = 2cos(5.0134
= 2(.29648...)
= appr .59297...
Right Side = -sin(2.5067)
= appr. -.593
≠ left side, so it does not work,
Try mine, it works, as shown in the graph.

Answer 6

Final word:

http://www.wolframalpha.com/input/?i=solve+2cos(2x)%3D-sinx

look at the graph, let your mouse hover over the red intersection points.

Answer 7

My correction:

x = 2.138 radians isn't correct solutions becouse:

2.13863 ≈ 2.139

Only x= 1.003 radians is correct solutions becouse:

1.00297 ≈ 1.003

Answer 8

2 - 4sin^2 x + sinx = 0
-4sin^2x + sin x + 2 = 0
a= -4
b= 1
c= 2

why are you multiplying by -1?

Answer 9

Because it is customary, and it often makes it easier, if the coefficient of the square term is positive.

So I switched all the signs, you must have known that this would not change the solution to the equation.

Answer 10

lol i see my mistake
i divided by 8 and not -8