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## Homework Statement

Solve acos²θ+bsinθ+c=0 for all values 0≤θ≤360°

a=16

b=6

c=-12

So 16cos²θ+6sinθ-12=0

## Homework Equations

Cos²x=1-Sin²x

## The Attempt at a Solution

Identity: Cos²x=1-Sin²x

16(1-Sin²θ)+6Sinθ-12=0

16-16Sin²θ+6Sinθ-12=0

6Sinθ-16Sin²θ=12-16=-4

Divide by 2(?)

3Sinθ-8Sin²θ=-2

3Sinθ-4Sin³θ=-2

3Sinθ-4Sin³θ=Sin3θ

Sin3θ=-2

I realised that I am stuck at this point and can’t work out how to find θ from this, although it feels like I am missing something obvious. Any help would be appreciated!