SOLUTION: Find to the nearest degree, all values of X in the interval 0°≤X≤360° which satisfy the equation 3sin^2X+2sinX-1=0
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Question 862112: Find to the nearest degree, all values of X in the interval 0°≤X≤360° which satisfy the equation 3sin^2X+2sinX-1=0 Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Find to the nearest degree, all values of X in the interval 0°≤X≤360° which satisfy the equation 3sin^2X+2sinX-1=0
(3sinx-1)(sinx+1)=0
..
3sinx-1=0
sinx=1/3
x=20˚, 161˚
..
sinx+1=0
sinx=-1
x=270˚
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