Hey guys,

Sorry I couldn't have a more direct title. There isn't really a name for this exact topic, but anyway, this is a rule we learnt:

asinθ +bcosθ = r sin (θ +λ)

where r =√a^2 + b^2

and tanSolve for 0°≤ θ ≤ 360°:λ= b/a

Question:

√2Cosθ + Sinθ = -1

My Working:r = √3

λ =54 degrees 44 minutes.

r sin (θ +λ) = -1

√3sin(θ + 54°44') = -1

sin(θ + 54°44') = -1/√3

θ + 54°44' = ?

I know you minus 54°44' from the answers you get. But from here onwards, I do not understand what to do. I've tried with answers but I'm failing.

Answers are 160°32', 270° for 0°≤ θ ≤ 360°.

Please help!

Many thanks.