How do you solve 5sin2(α)11sin(α)+2=0?

1 Answer
Aug 10, 2016

α=11.54

Explanation:

Let sin(α)=x
Hence we can write
5x211x+2=0
or
5x210xx+2=0
or
5x(x2)1(x2)=0
or
(x2)(5x1)=0
or
x2=0
or
x=2
or
5x1=0
or
5x=1
or
x=15
or
Out of the above two solutions sinα=2 which is invalid
Hence
sinα=15
or
α=sin1(15)
or
α=11.54