Find all real numbers on the interval [0, 2π) that satisfy the equation 15sin^2 = 4sinx + 3 [TEN POINTS!]?
Someone please show me how to do this and get the answer.
Find all real numbers on the interval [0, 2π) that satisfy the equation 15sin^2 = 4sinx + 3.
Thanks! i'll choose best answer [ten points] soon.
Update:But what are ALL of the specific answers? Each number that will work on this interval.
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Let Y = sin(x), then convert your equation into a quadratic in Y.
15sin(x)^2 = 4sin(x) + 3
15sin(x)^2 - 4sin(x) - 3 = 0
sin(x) = (4 +/- sqrt(16 + 180)) / 30
sin(x) = (4 +/- sqrt(196)) / 30
sin(x) = (4 +/- 14) / 30
sin(x) = (2 +/- 7) / 15
sin(x) = 9/15 , -5/15
sin(x) = 3/5 , -1/3
x = arcsin(3/5) , arcsin(-1/3)
15sin^2(x) - 4sinx - 3. = 0
(5sinx-3)(3sinx+1)=0
sinx = 3/5 or -1/3
x = arcsin (3/5) and arcsin(-1/3)