SOLUTION: Please help me find all solutions of the equation in the interval [0, 2pi)
2 sin^2x + 3 sinx + 1 = 0
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-> SOLUTION: Please help me find all solutions of the equation in the interval [0, 2pi)
2 sin^2x + 3 sinx + 1 = 0
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2 sin^2x + 3 sinx + 1 = 0
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let u=sin x
u^2=sin^2 x
2u^2+3u+1=0
(2u+1)(u+1)=0
2u+1=0
u=-1/2
or
u+1=0
u=-1
..
sin x=u=-1
x=π
..
sin x=-1/2
x=7π/6 and 11π/6 (In quadrants III and IV where sin is negative)
solutions:
x=π, 7π/6, 11π/6
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