answersLogoWhite

0


Best Answer

That equation is impossible to solve because Sin-1(3) is an impossible proportion in a triangle. Earlier in the problem you might have made a mistake. Or, if that is the whole problem you cannot deduce an answer because there is no value of X, Real or imaginary that works in that equation.

User Avatar

Wiki User

13y ago
This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: How would you solve x plus 3 equals sinx for x?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

Solve 6sinx equals 1 plus 9sinx algebraically over the domain 0 is greater than or equal to x is less than 2pi?

6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.


Sinx plus cosx equals 0?

x = 3pi/4


How do you solve 6sin x 1 plus 9sin x algebraically over the domain 0 x 2pi?

6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.


How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?

(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx


How would you solve sin3x equals sinx?

To start, try breaking down sin3x using a double angle formula. Message me if you need more help!


Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?

2


Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?

(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx


Solve 2sinx-sin3x equals 0?

2sinx - sin3x = 0 2sinx - 3sinx + 4sin3x = 0 4sin3x - sinx = 0 sinx(4sin2x - 1) = 0 sinx*(2sinx - 1)(2sinx + 1) = 0 so sinx = 0 or sinx = -1/2 or sinx = 1/2 It is not possible to go any further since the domain for x is not defined.


Is there any way to solve a system of equations with C and D as constants and x and y as variables sinx plus cozy - C equals 0 cosx plus siny - D equals 0?

0


How do you solve csc x-sin x equals cos x cot x?

cscx-sinx=(cosx)(cotx) 1/sinx-sinx=(cosx)(cosx/sinx) (1/sinx)-(sin^2x/sinx)=cos^2x/sinx cos^2x/sinx=cos^2x/sinx Therefore LS=RS You have to remember some trig identities when answering these questions. In this case, you need to recall that sin^2x+cos^2x=1. Also, always switch tanx cotx cscx secx in terms of sinx and cosx.


How do you solve 2sinxsinx equals 1?

2sinxsinx=1 (sinx)(sinx)=1/2 sinx=1/4=o.25 since, roughly, for small x values, sin x = x then x=0.25 Otherwise, to be more accurate, we proceed as follows: sinx=0.25 (as given before) then x=arc sin 0.25=0.2526803


Verify that Cos theta cot theta plus sin theta equals csc theta?

It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)