solve:
-sec= square root 2 for the interval 0<x<360
-
Presentation is open to doubt so will take a guess at :-
sec ∅ = √2
1 / cos ∅ = √2
cos ∅ = 1/√2
Ø = 45° , 315°
0<x< 360° same write 0<x<2π
-sec(x)=√2
Arcsec(-sec(x))=Arcsec(√2)
-(x)=Arcsec(√2)
x=-Arcsec(√2)
do you mean –sec θ = √2
solve for θ ??
sec θ = –√2 = 1/cos θ
cos θ = –1/√2
θ = 135º
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Answers & Comments
-
Presentation is open to doubt so will take a guess at :-
sec ∅ = √2
1 / cos ∅ = √2
cos ∅ = 1/√2
Ø = 45° , 315°
0<x< 360° same write 0<x<2π
-sec(x)=√2
Arcsec(-sec(x))=Arcsec(√2)
-(x)=Arcsec(√2)
x=-Arcsec(√2)
do you mean –sec θ = √2
solve for θ ??
sec θ = –√2 = 1/cos θ
cos θ = –1/√2
θ = 135º