A very good evening to you too, Mr Eric :)
Thank you so much for your advised solution. Let me try to digest the solution steps now....:))

Yes, I got it now:)
Once again thanks a lot, Mr Eric.

The other way to approach this :

243 + 135/2sin²x + 15/16sin⁴x = 405/2sinx + 45/4sin³x + 1/32sin^5x

Get rid of all denominators by multiplying the whole equation by sin^5x first.


243 sin^5 x + 135/2 sin³x + 15/16 sinx = 405/2 sin⁴x + 45/4 sin²x + 1/32

243 sin^5 x - 405/2 sin⁴x + 135/2 sin³x - 45/4 sin²x + 15/16 sin x - 1/32 = 0

(3sinx)^5 + (5)(-½)(3sinx)⁴ + (10)(-½)²(3sinx)³
+ (10)(-½)³(3sinx)² + (5)(-½)⁴(3sinx) + (-½)^5 = 0


This is now in the expanded form of the expression (3sinx - ½)^5 using binomial theorem. You should be able to see the Pascal's triangle pattern 1 5 10 10 5 1.


(3sinx - ½)^5 = 0

3sinx - ½ = 0

3sinx = ½

sinx = 1/6

basic angle θ = sin-¹ (1/6)

≈ 9.594° (3d.p)

x = 9.594° , 180° - 9.594°, 9.584° + 360°,...

Good morning Mr J, thank you for your alternative solution :)

Good morning Mr Eric and Mr J, I have just verbally confirmed with my school teacher that A maths technique can be used to solve E maths question and vice versa.

Thank you so much for your help rendered to me all these while. I am very grateful.

Have a good day ahead and pls take care :)

Great