Question

How do you solve `5sin^2x-4sinx-1=0` in the interval `0<=x<=2pi`?

Answer 1

`" "x=pi/2" "Or " "x=-0.201358`

Explanation:

Let `color(blue)(y=sinx)`
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`5sin^2 x - 4sinx = 1 =0`
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`rArr5y^2 - 4y - 1 =0`
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This equation is solved by factorizing `" "5y^2 - 4y - 1" "` and
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finding its quadratic roots.
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`delta = b^2 - 4ac = (-4)^2-4(5)(-1) = 16 + 20 = 36`
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`delta>0" "` Two roots exist in `RR`
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`y_1=(-b- sqrt(delta))/(2a)=(4-sqrt36)/(2xx5)=(4-6)/10=-2/10=-1/5`
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`y_2=(-b + sqrt delta)/(2a)=(4+sqrt36)/(2xx5)=(4+6)/10=10/10=1`
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The form of the equation by substituting the factorization form of the polynomial :
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`5y^2 - 4y - 1=0`
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`(y+1/5)(y-1)=0`
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`y+1/5=0rArry=-1/5rArrcolor(blue)(sinx=-1/5)rArrx=-0.2`
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`y-1=0rArry=1rArrcolor(blue)(sinx=1)rArrx=pi/2`
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Hence,`" "x=pi/2" "Or " "x=-0.201358`