0
The problem x = 2 sin x cannot be solved by using algebraic methods.
One solution is to draw the graphs of y = x and y = 2 sin x.
The two lines will intersect. The values of x where the intersection takes place are the solutions to this problem.
You can tell from the graph that one solution is x=0 and verify this contention by noting that 2 sin(0) = 0.
You can find the other solution through numerical methods and there are many that will give you the correct solution. Perhaps the simplest is to start with a value of X like pi/2 and then take the average of 2*sin(X) and X. Using that as your new value, again take the average of 2*sin(X) and X. As you continue to do this, the value will get closer and closer to the desired value. After 20 steps or so, the precision of your calculator will probably be reached and you will have a pretty good answer of about 1.89549426703398. (A spreadsheet can be used to make these calculations pretty easily.)
Still curious? Ask our experts.
Chat with our AI personalities
Rene
Fran
JordanAdd your answer:
Earn +20 pts
Determine the amplitude of y equals -2 sin x?
The amplitude of the wave [ y = -2 sin(x) ] is 2.
How do you solve sin 2x equals sin x over 2?
they do have calculators for these questions you knowsin 2x = (sin x)/22 sin x cos x - (1/2)sin x = 02 sin x(cos x - 1/4) = 02 sin x = 0 or cos x - 1/4 = 0sin x = 0 or cos x = 1/4in the interval [0, 360)sin x = 0, when x = 0, 180cos x = 1/4, when x = 75.52, 284.48Check:
Solution for tan x is equal to cos x?
Tan(x) = Sin(x) / Cos(x) Hence Sin(x) / Cos(x) = Cos(x) Sin(x) = Cos^(2)[x] Sin(x) = 1 - Sin^(2)[x] Sin^(2)[x] + Sin(x) - 1 = 0 It is now in Quadratic form to solve for Sin(x) Sin(x) = { -1 +/-sqrt[1^(2) - 4(1)(-1)]} / 2(1) Sin(x) = { -1 +/-sqrt[5[} / 2 Sin(x) = {-1 +/-2.236067978... ] / 2 Sin(x) = -3.236067978...] / 2 Sin(x) = -1.61803.... ( This is unresolved as Sine values can only range from '1' to '-1') & Sin(x) = 1.236067978... / 2 Sin(x) = 0.618033989... x = Sin^(-1) [ 0.618033989...] x = 38.17270765.... degrees.
Is sin 2x equals 2 sin x cos x an identity?
YES!!!! Sin(2x) = Sin(x+x') Sin(x+x') = SinxCosx' + CosxSinx' I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value. Hence SinxCosx' + CosxSinx' = Sinx Cos x + Sinx'Cosx => 2SinxCosx
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
