Solving a trigonometric equation with sixth powers

[@gustavomedeiros1669]

You can also use the fact that sin2x = 2sinxcosx so sin^2xcos^2 = (sin^2x)/4 to avoid the substituition. Also all the solutions can be written as +-pi/6 +npi/2

[@SyberMath]

Nice

[@p12psicop]

u^2-u-3/16

[@damiennortier8942]

(a+b)^3 = a^3 + b^3 + 3ab(a + b) and sin^2x + cos^2x = 1 So you got sin^6x + cos^6x = 3sin^2xcos^2x - 1 And then, sin^2xcos^2x = - 3/16. Let sin^2x = a and cos^2x = b and you got a system : a^3 + b^3 = 7/16 and ab = - 3/16 So we can find a+b : (a+b)^3 = 7/16 - (a+b)/16. Call a+b, u to got a cubic : 16u^3 + u - 7 = 0 Let factor by 16(u - a)(u^2 + bu + c). Since this give us 16u^3 + 16(b - a)u^2 + 16(c - ab)u - 16ac, we got a system : 16(b - a) = 0 16(c - ab) = 1 16ac = 7 ...

[@petermichaelgreen]

My thoughts were to write it in terms of (cos²x)³ and (sin²x)³ then replace cos²x with 1-sin²x. After multiplying it out and simplifying the (sin²x)³ terms will cancel leaving us with a quadratic in sin²x. Bung that into the quadratic formula, square root it and finish off with the inverse sine.

[@angelmendez-rivera351]

This was pleasantly comprehensive.

[@SyberMath]

Thank you!

[@ManjulaMathew-wb3zn]

The general solution can be stated in one line as (3n+/-1)PI/6.

[@aashsyed1277]

u a pro

[@artsmith1347]

There were four solutions at the end of the video. A 6th degree equation has six roots. Are two of them repeated in the solutions at the end of the video?

[@petermichaelgreen]

If you make the substitution cos²x = 1-sin²x to replace sin with cos or cos with sin in the original equation you will find the 6th degree terms cancel and there are no 5th degree terms. So it's actually only a fourth degree equation.