Hard Trigonometry Question: Solve cos 3x + sin 2 x = 1?

With t=tan(x/2)

cos(3x) = 4cos³x−3cosx = 4(1−t²)³/(1+t²)³ − 3(1−t²)/(1+t²)

sin(2x) = 2 * 2t/(1+t²) * (1−t²)/(1+t²)

∴ 4(1−t²)³ − 3(1−t²)(1+t²)² + 4t(1−t²)(1+t²) = (1+t²)³

→ t(t⁵+2t⁴−6t³+9t−2) = 0

t=0 gives x=0,2π

t⁵+2t⁴−6t³+9t−2 = 0

Has no simple linear or quadratic factors so requires numerical method.

tan(x/2) ≈ −3.49308 → x = 3.69924

tan(x/2) ≈ −1.23653 → x = 4.50166

tan(x/2) ≈ 0.229603 → x = 0.451383

Note about http://uk.answers.yahoo.com/question/index?qid=201...

I would have been more than happy to have given just an algebraic solution if that’s what you wanted but

since the question offered specific values I assumed you wanted a specific answer.

Implementing r = c + (p−c•n)n + URcosθ + VRsinθ (U & V orthonormal) in this case requires undergoing roughly the same calculations as I did. However, this formula assumes that the intersection is a circle and that the centre is at c + (p−c•n)n which makes the calcs a bit easier. My answer first showed the intersection was a circle by deriving the familiar Cartesian equation, identified the centre and then applied a parameterization.

Thanks for some interesting questions.