2 sin x = cos(x/3)
This is a pretty tough one.
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Let's start by setting y=x/3 so x=3y and substituting. Then we can use the triple angle formula:
2 sin(3y) = cos y
2 (3 sin y - 4 sin ^3 y ) = cos y
Let's square so we write everything in terms of sin^2 y. This will likely introduce extraneous roots.
4 sin ^2y (3 - 4 sin ^2y)^2 = cos ^2 y = 1 - sin^2 y
Let s= sin ^2 y. Squared sines are called spreads in Rational Trigonometry.
4 s(3 - 4s)^2 = 1 - s
4 s( 9 - 24 s + 16 s^2) = 1 - s
64 s^3 - 96 s^2 + 37 s - 1 = 0
That's a cubic equation with three real roots, candidates for the squared sines of 3x. We could employ the cubic formula, but that will just lead to some cube roots of complex numbers that aren't particularly helpful. Let's just take a numerical solution :
s≈0.66035 or s≈0.029196 or s≈0.81045
x = 3y = 3 arcsin (pm sqrt{s})
Let's work in degrees. Our potential approximate solutions are:
x = 3 arcsin(\pm sqrt{ 0.66035 } ) approx \pm 163.058^circ or \pm 703.058^circ
x = 3 arcsin(\pm sqrt{ 0.029196 } ) approx \pm 29.5149^circ or \pm 569.51^circ
x = 3 arcsin(\pm sqrt{ 0.81045 } ) approx \pm 192.573^circ or pm 732.573^circ
Let's see if any of those work. Let e(x)=2 sin x - cos ( x/3)
e( 163.058^circ) approx 0.00001 quad that's a solution.
e(-163.058^circ) approx -1.17 quad not a solution.
Clearly at most one of a \pm pair will work.
Ten more to go.
e( 703.058^circ ) approx 0.00001 quad sqrt
e( -703.058 ^circ) quad nope
e( 29.5149^circ ) approx 10^{-6} quad sqrt
e( -29.5149^circ ) quad nope
e( 569.51^circ ) approx 10^{-4} quad sqrt
e( -569.51^circ ) quad nope
e( 192.573^circ ) approx -.87 quad nope
e( -192.573^circ ) approx 0.00001 quad sqrt
e( 732.573^circ ) approx -.87 quad nope
e( -732.573^circ ) approx 0.00001 quad sqrt
The arcsin comes with a + 360^circ k, and the factor of three makes it 1080^circ k.
OK, our approximate solutions are:
x = { 163.058^circ, 703.058^circ, 29.5149^circ, 569.51^circ , -192.573^circ , -732.573^circ } + 1080^circ k quad for integer k.
