Find the value of x :: cos^3 x.sin 3x+sin^3 x.cos 3x=3/4?

Question

[Cos cube x] is denoted by [cos^3 x].
Please solve it. I will give you 10 points.

Answer 1

cos^3x(3sinx-4sin^3x)+sin^3x(4cos^3x-3cosx)=3/4
3sinxcos^3x-4sin^3xcos^3x+4sin^xc0s^3x-3sin^3xcosx=3/4
3sinxcosx(cos^2x-sin^2x)=3/4
3/2(sin2xcos2x)=3/4
3/4sin4x=3/4
sin4x=sinpi/2

using sin theta=sin alpha
the general solution is 4x=n*pi+(-1)^n*pi/2
so x=n*pi/4+(-1)^n*pi/8 where nEZ

Answer 2

subtract 3x from the 5x and upload 7 to the three so variables and numbers are at the same time on opposite factors so : 3x+5=5x-7 -3x -3x +7 +7 makes it: 12=2x then divide the two factors by utilising 2 so x is on my own making x=6

Answer 3

cos^3x.sin3x + sin^3x.cos3x
= cos^3x(3sin x - 4sin^3x) + sin^3x(4cos^3x - 3cos x)
= cos^3x.sin x(3 - 4sin^2x) + sin^3x.cos x(4cos^2x - 3)
= sin x.cos x(3cos^2x - 4sin^2x.cos^2x + 4cos^2x.sin^2x -3sin^2x)
= 3sin x.cos x.cos2x
= 3/2(sin2x.cos2x) [divide , multiply by 2].

Now,
3/2(sin2x.cos2x) = 3/4
=> sin2x.cos2x = 1/2
Recall that sin x and cos x are positive in quadrant 1 and both negative in 3.
As 1/2 is +ve, 2x has to lie in 1 or 3.
For sin2x.cos2x = 1/2 in quad 1,
sin2x = cos2x = 1/root of 2
=> 2x = pii/4 or 45degrees.
=> x = pii/8
Similarly in quad 4,
2x = 5 pii/4
=> x = 5pii/8

THEREFORE, x = pii/8 , 5pii/8.

(hope u know the formulae of sin3x and cos3x.)

Answer 4

I will not do your home work but I will help you with it.

If you divide both sides of the equation by sin 3x+sin^3. convert either cos to sin or sin to cos.

You you guess witch one will be easier, both will work.

Now it is up to you.

I do not need ten points but you do need to do the home work(I am guessing it is your home work).

Answer 5

you're weird!