Question 972381: Solve the following equations for 0 < (x) < 360
Sin^3(x)+sin^2(x)=0
sin^2(x)+4sin(x)-5=0
Notes inconclusive and not able to find clear info online
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Solve the following equations for 0 ≤ (x) < 360
Sin^3(x)+sin^2(x)=0
sin^2(x)+4sin(x)-5=0
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Sin^3(x)+sin^2(x)=0
sin^2(x)(sinx+1)=0
..
sin^2(x)=0
sinx=0
x=0, 180˚
or
sinx+1=0
sinx=-1
x=270˚
..
sin^2(x)+4sin(x)-5=0
(sinx+5)(sinx-1)=0
sinx=-5 (reject,(-1 < sinx < 1))
or
sinx=1
x=90˚
..
Note: I assumed you meant given domain is 0 ≤ (x) < 360 instead of 0 < (x) < 360, otherwise, there is no solution to sinx=0, since 0 is not in the domain
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