Solve for x E [0,2pi]: sin(3x) - sin(x) = cos(2x)
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
Hello, Mr_Green!
We need a sum-to-product identity: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