Solve for x E [0,2pi]: sin(3x) - sin(x) = cos(2x)

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- April 10th 2007, 01:23 PMMr_Greentrig: find x's
Solve for x E [0,2pi]: sin(3x) - sin(x) = cos(2x)

- April 10th 2007, 02:06 PMJhevon
let's expand this

sin(3x) - sin(x) = cos(2x)

=> sin(2x)cos(x) + sin(x)cos(2x) - sin(x) = cos^2(x) - sin^2(x)

=> [2sin(x)cos(x)]cos(x) + sin(x)[cos^2(x) - sin^2(x)] - sin(x) = cos^2(x) - sin^2(x)

=> 2sin(x)cos^2(x) + sin(x)[cos^2(x) - sin^2(x)] - sin(x) = cos^2(x) - sin^2(x)

we have a lone sin(x), so let's change everything to sin(x)

=> 2sin(x)(1 - sin^2(x)) + sin(x)[1 - sin^2(x) - sin^2(x)] - sin(x) = 1 - sin^2(x) - sin^2(x)

=> 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) - sin(x) = 1 - 2sin^2(x)

=> 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) - sin(x) - 1 + 2sin^2(x) = 0

now this is ugly, let's try and simplify this a bit

=> -4sin^3(x) + 2sin^2(x) + 2sin(x) - 1 = 0

this is a cubic in sin(x), to cut down on typing for the mean time, let's write sin(x) as y, we get:

-4y^3 + 2y^2 + 2y - 1 = 0 ........ah, that looks a little better

=> -2y^2(2y - 1) + (2y - 1) = 0

=> (2y - 1)(1 - 2y^2) = 0

=> y = 1/2 or y = +/- sqrt(1/2)

but, y = sin(x)

=> sin(x) = 1/2 or sin(x) = +/- sqrt(1/2) = +/- 1/sqrt(2) = +/- sqrt(2)/2

=> x = pi/6, 5pi/6 or x = pi/4, 3pi/4, 5pi/4, 7pi/4

so there are 6 values of x for which this happens - April 10th 2007, 02:08 PMJhevon
hmm, it seems i've made an error somewhere. only two of my values are actual solutions (i graphed the functions to see)

there are 6 values though. pi/6 and 5pi/6 work - April 10th 2007, 02:33 PMSoroban
Hello, Mr_Green!

Quote:

Solve for x ε [0,2π]: .sin(3x) - sin(x) .= .cos(2x)

. . sin(A) - sin(B) .= .2·cos(½(A + B)]·sin[½(A - B)]

The left side is:

. . sin(3x) - sin(x) .= .2·cos[½(3x + x)]·sin[½(3x-x)] .= .2·cos(2x)·sin(x)

The equation becomes: .2·cos(2x)·sin(x) .= .cos(2x)

. . . . . . . . . . .2·cos(2x)·sin(x) - cos(2x) .= .0

Factor: .cos(2x)·[2·sin(x) - 1] .= .0

We have two equations to solve:

cos(2x) = 0 . → . 2x = π/2, 3π/2, 5π/2, 7π/2 . → . x = π/4, 3π/4, 5π/4, 7π/4

2·sin(x) - 1 .= .0 . → . sin(x) = ½ . → . x = π/6, 5π/6

- April 10th 2007, 02:33 PMMr_Green
don't worry about it. i will look through your work, and let you know if I find anything.

thanks - April 10th 2007, 02:47 PMMr_Green
Jhevon I checked your work and everything is in order. i also graphed them and they were all correct.