Question

How to solve these equations?

Answer 1

1) / sin x / = y ≥ 0 ;
We solve the equation
`y^(2) +y - 2 = 0 ;`
a = 1 ; b = 1 ; c = - 2 ;

`b^(2) - 4ac = 9 ;`
y1 = ( - 1 + 3 ) / 2 = 1 ≥ 0 ( correct ) ;
y2 = ( - 1 - 3 ) / 2 = - 2 ≥ 0 ( false ) ;
Then, / sin x / = 1 ;
sinx = + 1 or sin x = - 1 ;
x ∈ { 90 } U { 270 } ;
x ∈ {90 , 270}.