Find all solutions of the given equation that lie on the interval [0, pi]:
8sin^2(x)cos^2(x)=1.
I have tried using the double-angle formulas, the half-angle formulas, and the fundamental identity of trigonometry. I keep getting stuck. (This is not for homework. I do math problems for fun!) Please show and explain your work so that I can figure out how to do other problems like this. Thank you.
8sin^2(x)cos^2(x)=1.
I have tried using the double-angle formulas, the half-angle formulas, and the fundamental identity of trigonometry. I keep getting stuck. (This is not for homework. I do math problems for fun!) Please show and explain your work so that I can figure out how to do other problems like this. Thank you.
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8 * sin(x)^2 * cos(x)^2 = 1
2 * 4 * sin(x)^2 * cos(x)^2 = 1
4 * sin(x)^2 * cos(x)^2 = 1/2
(2 * sin(x) * cos(x))^2 = 1/2
sin(2x)^2 = 1/2
sin(2x) = +/- sqrt(2) / 2
2x = pi/4 + pi/2 * k
2x = (pi/4) * (1 + 2k)
x = (pi/8) * (1 + 2k)
x = pi/8 , 3pi/8 , 5pi/8 , 7pi/8
2 * 4 * sin(x)^2 * cos(x)^2 = 1
4 * sin(x)^2 * cos(x)^2 = 1/2
(2 * sin(x) * cos(x))^2 = 1/2
sin(2x)^2 = 1/2
sin(2x) = +/- sqrt(2) / 2
2x = pi/4 + pi/2 * k
2x = (pi/4) * (1 + 2k)
x = (pi/8) * (1 + 2k)
x = pi/8 , 3pi/8 , 5pi/8 , 7pi/8
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2 sin^2 (2x) = 1 ---> sin(2x) = +/- 1/sqrt2 , then easy to solve for x