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prabhleenkaur~:
Hi!

Could I please get help on this question?

sin2x + sin3x + sinx =0
using general solutions

Thank you.

kiwiberry:

--- Quote from: prabhleenkaur~ on January 22, 2017, 04:47:50 pm ---Hi!

Could I please get help on this question?

sin2x + sin3x + sinx =0
using general solutions

Thank you.

--- End quote ---

sinx + 2sinxcosx + 3sinx - 4sin3x = 0
2sinx(2 + cosx - 2sin2x) = 0
2sinx[-2(1-cos2x) + cosx + 2) = 0
sinx(2cos2x + cosx) = 0
sinxcosx(2cosx +1) = 0
therefore sinx=0, cosx=0 or cosx=-1/2

sinx = 0 = sin0
∴ x = πn

cosx = 0 = cos(π/2)
∴ x = 2πn ± π/2

cosx = -1/2 = cos(2π/3)
∴ x = 2πn ± 2π/3

jamonwindeyer:
Thanks kiwiberry, you are a legend!! ;D prabhleenkaur~, welcome to the forums! ;D

hanaacdr:
Hi
i have a similar question,
would i be able to get some help on this,

sin2x + sin3x + sin4x = 0

much appreciated thank you!

jakesilove:

--- Quote from: hanaacdr on January 23, 2017, 04:01:20 pm ---Hi
i have a similar question,
would i be able to get some help on this,

sin2x + sin3x + sin4x = 0

much appreciated thank you!

--- End quote ---

For a question like this, I would expand out into just sin(x) and cos(x)







Well, this was a bad idea. I'll leave this here anyway.

Let's break it down into sin(2x)?





Nup. Did you type out the question correctly? If I wolfram alpha the solution, it's pretty damn complicated.

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