Question 459274
First we simplify this equation:{{{4(1-(sinx)^2)-5sinx-5=0}}}=>

=>{{{4-4(sinx)^2-5sinx-5=0}}}=>{{{4(sinx)^2+5sinx+1=0}}}. 

In the last equation we substitute {{{sinx=y}}}, and get the quadratic equation:

{{{4y^2+5y+1=0}}}, solving this equation we get: y=-1 and y=-1/4.

Remembering our substitution {{{y= sinx}}} we need to solve two simple 

trigonometric equations equivalent with our original equation.

1){{{sinx=-1}}}, where {{{x=3*pi/2}}}, and

2){{{sinx=-1/4}}}, where {{{x=-0.25}}}, and {{{x=2*pi+0.25}}}.