is this possible..??

Question

I've managed to pull and equation this far... I'm not sure if its possible to finish...

48cosX = 0.4(980 + 48sinX)

x being an unknown angle, of course....can anyone solve this?

I've decided this is NOT possible, but thnx for the help anways =)

Answer 1

I am not sure if there is a solution. I hope I am mistaken.

48cos(x)=0.4(980+48sin(x))
120cos(x)=980+48sin(x)
120cos(x)-48sin(x)=980

But since sin(x) and cos(x) can never exceed 1, so the above equation cannot have a solution, since 980>|120|+|-48|

Answer 2

Write it as
48 cos x - 0.4*48 sin x = 0.4*980
i.e.
48 cos x - 19.2 sin x = 392

Now since
cos (x + a) = cos x cos a - sin x sin a,
the LHS would be very nice if 48 were equal to cos a and 19.2 were equal to sin a. This isn't possible, but if we divide both sides by sqrt(48^2 + 19.2^2) we get
0.928476691*cos x - 0.371390676*sin x = 7.582559642

Now the left side is equal to cos (x + a), where
cos a = 0.928476691
and sin a = 0.371390676

So a = 21.80140949 degrees.

That means your equation has no solutions, because cos(x+a) must be less than 1, and the RHS is greater than 1.

If it had been a value between -1 and 1, you could grab your calculator, do an inverse cosine to get the value of x + a in degrees, and then subtract 21.80140949.

Answer 3

This is a transcendal equation. The only way to solve it and find out at the same time if it is solveable at all is to graph both sides; where they meet is the solution set of x; there may be more than one value of x that satisfies both sides of the equation.

A numerical solution is possible by iteration: solve left side by dividing both sides by 48, then you get

cosX = 0.4(980 + 48sinX)/48 or
X=arccos[0.4(980+48sinX)/48]

start by letting x =1 on the RHS, then use the value you get for X on the LHS into X on the RHS and keep doing this until you get some convergence.

Graphical solutions are much easier especially if you have one of those graphing calculators.

Answer 4

No, it's NOT possible!!! :)

If you do the algebra to clean up the equation, you get:
cos(x) - 0.4sin(x) = 8.1666...
The range of the sine and cosine functions is -1 to 1. Therefore, the maximum value that cos(x) - 0.4sin(x) can POSSIBLY have is 1 - 0.4(-1) = 1.4, which is a lot less than 8.1666.

Answer 5

48cosX = 0.4(980 + 48sinX)
48cosX=392+.4*48sinX
48(cosX-.4sinX)=392
cosX-.4sinX=8.16666

since the max value of cos x is 1 & the min value of sin x is -1
and 1-.4*(-1)=1.4

there is no solution to this problem ie there is no value of X that makes this true.

Answer 6

I would square both sides, then replace cos^2(x) with 1-sin^2(x) on the left side. That would leave a quadratic equation for sin(x) which is easy to solve.

Good luck!

Edit: Good answer just below mine - this means that my quadratic equation would have solutions (if any) outside of [-1,1] for sin(x).

Answer 7

this is just plan horrible
maybe i am forgetting a special trig function but this cant be right if you need to solve this.
that farthest i got was cos(x)-.4(sin(x)+20.42)=0
i dont think you can go further.

Answer 8

eww thats such a horrible looking equation
give us the original question, because maybe youve gone wrong somewhere.