A levels maths query

[verbal linden]

Supposedly 104.5 is extra. Please explain why?
I have added the mark scheme for reference

[fossil bramble]

Does plugging it into the original equation gives you 3?

[molten knoll]

You substituted in cos(θ) = √(1-sin^2(θ)), which doesn't take account into case where cos(θ) is negative, but then squared it, which makes the negative result of cos(θ) positive.

It would be easier if you cancelled out sin θ in earlier steps (also stating it only works if sin θ ≠ 0)

[visual iris]

So in the step where u converted cos into sqrt (1-cos^2)
That actually gave u extra solutions
Because u see sqrt of any number say x is equal to |x| which means ±x it gives u negative solutions too

U can pull out sin as common factor in the eqn and continue...that's the hint
Feel free to ask again if u r stuck

[molten knoll]

That doesn't make sense. '√' is a notation for principal square root so √(1) is 1. I can't comprehend the third sentence of your comment: I assume you don't mean √(x) = |x| = ±x for all(?) x, but I'm comprehending your phrasing like that?

[halcyon wedge]

You can Start by distributing the -3 into the parentheses
tan θ + 2 sin θ - 3 tan θ + 6 sin θ = 0 then substitute the tan since tan θ = sin θ/cos θ
Multiply by cos θ

(assuming cos 𝑄 ≠ 0)

[verbal linden]

Can u pls elaborate ?

I mean sqrt(1-sin²x) is cosx so wht ever i do with sqrt(1-sin²x) shldnt change the fact that it is cosx and hence shldnt give me extra answers?

Ts is bugging me for the past few hours 😭😭

[molten knoll]

Is your claim that cos x = √(1-sin^2(x)) for all real x?

I disprove that by counter-example: Substitute in x = 3π/4 into cos x leads to cos(3π/4) = -√(2)/2 but √(1-sin^2(3π/4)) = √(1-1/2) = √(1/2) = 1/√(2) = √(2)/2 ≠ -√(2)/2. QED.
So you can't substitute cos x = √(1-sin^2(x)) into the equation if you're only given that x is a real number.

[verbal linden]

Ohhh i see

Makes sense

Thank you!