What is the intersection of 5sinx and 3cos2x?

Question

I've been working on this for a few days.

r(x)=5sinx, s(x)=3cos2x

both have the domain of [0, 2pi).

I tried to make the two functions equal to each other and I got so far as :

5sinx=3cos2x

5sinx=3(1-2sin²x)

(5/3)sinx=1-2sin²x

(5/3)sinx=(1-sin²x) - sin²x

(5/3)sinx=cos²x - sin²x

I am probably doing this wrong... any tips and next step suggestions, please?

Answer 1

5 sinx - 3 cos 2x = 0

5 sin x - 3 (1 - 2 sin ² x) = 0
5 sin x - 3 + 6 sin ² x = 0
6 sin ² x + 5 sin x - 3 = 0
sin x = [- 5 ± √(25 + 72)] / 12
sin x = [- 5 ± √(97)] / 12
sin x = 4.85 / 12 , sin x = - 14.8 / 12
sin x = 0.404 , sin x = - 1.23
Accept sin x = 0 . 404
x = 23.8°
y = 5 sinx = 5 x 0.404 = 2.02
Intersect at (23.8° , 2.02)

Answer 2

I think you almost had it here ...

5sinx=3(1-2sin²x)

Distribute the "3", and set the thing = 0

6sin²x + 5sinx - 3 = 0

This is in "quadratic form", soved with "y" substitution.
Let y = sinx.
Then we have 6y² + 5y - 3 = 0
which factors to (3y - 1)(2y + 3) = 0
Solutions, y = 1/3 and y = -3/2

So, sinx = 1/3 or sinx = -3/2

Sove with inverse sines, or arcsine, or whatever your professor calls it.

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This is a common technique, to rewrite an equation entirely in sines or cosines, and get the problem in "quadratic form".

Answer 3

using the double perspective identity for cos cos 2x = a million-2sin^2 x 3(a million-2sin^2 x)=5sinx +4 3 - 6sin^2 x = 5sin x + 4 6sin^2 x + 5 sin x + a million = 0 enable y =sin x 6y^2 + 5y + a million = 0 (3y + a million) (2y + a million) = 0 y= -a million/3, y= -a million/2 sin x = (-a million/3), sin x = -a million/2 x= Arcsin (-a million/3), x= -pi/6 or 11pi/6

Answer 4

Go back to
5 sin x = 3 (1 - 2 sin² x)
=> 5 sin x = 3 - 6 sin² x
=> 6 sin² x + 5 sin x - 3 = 0
=> sin x = (-5 ± √(25 - 4(6)(-3))) / 12
=> sin x = (-5 ± √97) / 12
However, (-5 - √97) / 12 < -1, so we must have
sin x = (-5 + √97) / 12 ≈ 0.40407.
=> x = arcsin((-5 + √97) / 12 ), π - arcsin((-5 + √97) / 12 )
= 0.41596, 2.72569 (5 d.p.)

You can check that both solutions work.

Answer 5

5 sinx = 3 cos2x
let u = sinx
cos2x = cos^2x - sin^2x = 1 - 2sin^2x = 1 - 2u^2
so
5u = 3(1- 2u^2) = 3 - 6u^2
or
6u^2 + 5u - 3 = 0
u = [ -5 +or- sqrt(25 - 4(6)(-3) ] /12
= [ -5 +or- sqrt(97) ] /12
= [ -5 +or- 9.85 ] /12
= 4.85/12, the only answer within -1 and +1
= 0.404
sinx = 0.404
x = 23.84 deg
or 180-23.84 = 156.16 deg

Answer 6

Your last step:

(5/3)sinx=cos²x - sin²x

Next step:

(5/3)sinx=cos²x + sin²x - 2*sin²x
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It's obvious there are 2 solutions:
http://s210.photobucket.com/albums/bb64/oregfiu/?action=view¤t=sinus.jpg
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