Solving an octic trigonometric equation

[@aashsyed1277]

OH MY GOD!!!!!!!!!!!!!!!! THE BEST CHANNEL!!!!!!!!!!!

[@aliscais6033]

Ihave a bit another solution with substitutions: (sinx)^2 = a; (cosx)^2 = b Then we have system of two equations: a^4 + b^4 = 97/128, a + b = 1 (a + b)^4 = a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 = a^4 +b^4 + ab(4*(a+b)^2 - 2ab) 1 = 97/128 + 4ab - 2*(ab)^2 ab = y 1 = 97/128 + 4y - 2*y^2 2*y^2 - 4y + 2 = 1 + 97/128 = 225/128 (y - 1)^2 = 225/256 = (15/16)^2 y = 31/16 or y = 1/16 y = ab = (sinx)^2 * (cosx)^2 = 1/4 * (sin2x)^2 (sin2x)^2 = (31)^(1/2)/4 > 1 => no solution (sin2x)^2 = 1/4 => sin2x = +-1/2 2x = +- pi/6 + pi*n x = +- pi/12 + pi*n/2

[@honestadministrator]

From 4-53 mnt onwards another way be used (2 cos(2x)) ^4 + 24 (2 cos(2x)) ^2 - 81 = 0 (( 2 cos(2x)) ^2 + 27 ) ( (2 cos(2x)) ^2 -3 ) = 0 or cos (2x) = √3/2, - √3/2, or 2x = n*π + π/6, n*π - π/6, or x = n*π /2 + π/12, n*π/2 - π/12 Hereby cos(2x)

[@adailtonjunior3736]

i found 4 solutions , and wolfram alpha find the same solutions and more one solution! I solved this question this way: (sen^2+cos^2)^4=sen^8+cos^8+2(sencos)^2(2(sen^4+cos^4)+3(sencos)^2) and this is equal to 1 because sen^2+cos^2 is equal to 1 than i supstitue (sencos)^2 with p and found that sen^4+cos^4 is equal to 1-2p. Then i put this in the first equation and found two values to p, that makes we found 4 values to (sencos)^2 and with this values we find 8 values to sen+cos, but some of them we dont count Than, i supstitute cos(x) with sen(pi/2 - x) to get sen(x)+ sen(pi/2 - x) that is equal to sqrt(2)(cos(x-pi/4), and finally we set this to be equal to the values we found before, findind 4 solutions for x i dont think you will read this, but i really appreciate your videos and sorry for my bad english, thats it!

[@voltalimwabbit5492]

The solutions are x = 15, 75, 105 and 165 if we use a quadratic equation where u = cos^2 2x and we will find x = 1/2 sqrt3 and -1/2 sqrt3

[@ashishpradhan9606]

It's a nice approach!!!

[@SyberMath]

Thank you! 😊

[@user-ez1in7rq6w]

When you got to cos⁴2x + 6cos²2x + 1 = 97/16 you could've just supstitute cos²2x with t and solve a quadratic equation

[@SyberMath]

That's right!

[@KevinAPamwar]

nice.... you used Cos(2X)

[@cameronspalding9792]

@ 4:54 I would have chosen to let u=cos^2(2x) and turn it into a quradratic

[@stu007]

Another question please : if cos^10(x) + sin^10(x)=11/36, find cos^12x+sin^12x

[@yapsiauwsoengie6507]

13/54

[@hkemal2743]

To the point!

[@SyberMath]

Thanks!

[@barakathaider6333]

👍👍

[@ИванИванов-я4л1в]

Hello! I have solved it in a different way

[@SyberMath]

How?

[@ИванИванов-я4л1в]

@@SyberMath I do not know why but my answers are deleted

[@VerSalieri]

Is the comment section too small for the solution?

[@AllanPoeLover]

2：40看不懂了

[@yapsiauwsoengie6507]

4a³b and 4ab³ will be eliminated when (a-b)⁴+(a+b)⁴ expanded

[@theuserings]

in (a-b)⁴, 4a³b and 4ab³ will be the negative terms, so when added with (a+b)⁴, where all the terms are positive, 4a³b and 4ab³ terms will be eliminated