Further Trig Question (1 Viewer)

[bujolover]

Hi guys,
I have a strong feeling that the given solution to the following question incorrectly missed out on α = 360°. Please confirm this for me.

Solve 12sin³α + 2cos3α - 9sinα + sin3α = 2 (0°≤α≤360°), using the fact that 12sin³α + 2cos3α - 9sinα + sin3α = 2√2[cos(3α + 45°)].

I got α = 0°, 90°, 120°, 210°, 240°, 330° or 360°.

Thanks in advance!

 

[1729]

Hi guys,
I have a strong feeling that the given solution to the following question incorrectly missed out on α = 360°. Please confirm this for me.

Solve 12sin³α + 2cos3α - 9sinα + sin3α = 2 (0°≤α≤360°), using the fact that 12sin³α + 2cos3α - 9sinα + sin3α = 2√2[cos(3α + 45°)].

I got α = 0°, 90°, 120°, 210°, 240°, 330° or 360°.

Thanks in advance!

360° is correct just make sure the inequality 0°≤α≤360° was not strict.

 

[InteGrand]

Hi guys,
I have a strong feeling that the given solution to the following question incorrectly missed out on α = 360°. Please confirm this for me.

Solve 12sin³α + 2cos3α - 9sinα + sin3α = 2 (0°≤α≤360°), using the fact that 12sin³α + 2cos3α - 9sinα + sin3α = 2√2[cos(3α + 45°)].

I got α = 0°, 90°, 120°, 210°, 240°, 330° or 360°.

Thanks in advance!

 

[bujolover]

It did.

 

[bujolover]

360° is correct just make sure the inequality 0°≤α≤360° was not strict.

What do you mean by the bolded part?

 

[InteGrand]

What do you mean by the bolded part?

 

[bujolover]

Oh no, it was ≤.