Question

Given that 0≤β<2π, find all the solutions to the equation 2sin^2β=7sinβ+4.

Answer 1

Explanation:

2sin²β-7sinβ-4=0

2sin²β-8sin β+1 sin β-4=0

2 sin β(sin β-4)+1(sin β-4)=0

(sin β-4)(2sin β+1)=0

either sin β-4=0

sin β=4 (rejected,as |sin β|≤1)

or

2 sin β+1=0

sin β=-1/2=-sin (π/6)=sin (π+π/6),sin (2π-π/6)

β=7π/6,11π/6