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3 votes
How to solve these equations?


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User Mcloving
by
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1 Answer

4 votes
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}.
User Sahu
by
9.1k points