3-sinx=cos(2x)?

Question

solve for x in radians, between 0 and 2pi

Answer 1

Using cos(2x)=cos^2x-sin^2x, then cos^2x=1-sin^x we can
rewrite your original equation as
3-sinx=-sin^2x-sin^2x+1 and combining we get

2sin^2x - sinx +2 =0. Now let sin x = y and rewrite as
2y^2 -y +2 =0 and apply the quadratic formula we get
y={1+/-(-15)^1/2}/4. But with a -15 in the radical we get
imaginary values for y so something is wrong with your
original equation.

Answer 2

There are no solutions.

Look at it this way:

(3-sinx) - cos(2x) = 0

The minimum value for (3 - sin x) is 2, and the maximum value for cos(2x) is 1, so (3-sinx) - cos(2x) is always greater than 0.

Answer 3

Simply speaking,
3 - sinx = cos2x ->
3 = cos2x + sinx ->

but max value for cos 2x and sinx can be 1. Hence max value for cos2x + sinx is 2. It can nevere take the value of 3.

Answer 4

<=> 3 - six = 1- 2 sin^2 x
let t be sinx , |t| < =1
<=> 2t^2 - t +2 =0
delta = b^2 - 4ac = -7 <0
so this equation has no root

Answer 5

Not possible.
Moving the sine over, you get 3 = sin(x) + cos(2x).
Howver, the maximum value for the function either sine or cos is 2.

Answer 6

3-sin x=cos2x
3-sin x=1-2sin^2x
2sin^2x-sin x+2=0
sin x={1+-sqrt[1-16]}/4
=1/4+-1/4*[sqrt-15]
There is something wrong
in the given equation.Sin x
can never have imaginary values.

Answer 7

No solution ... sin and cos never get larger than one or less than one.

Answer 8

Really?

Answer 9

impossible

Answer 10

No chance!!