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Alize Kuhn

Feb 20, 2025

sin4x-cos2x=0 How do you solve?

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Lavinia Carroll

Feb 20, 2025

sin(4x) - cos(2x) = 02sin(2x)cos(2x) - cos(2x) = 0cos(2x)(2sin(2x) - 1) = 0cos(2x) = 0 in which case 2x = pi/2, 3pi/2, 5pi/2x = pi/4, 3pi/4, 5pi/4x = (pi/4 + npi/2) where n is an integeror 2sin(2x) - 1 = 02sin(2x) = 1sin(2x) = 1/22x = pi/6, 5pi/6, 13pi/6, 17pi/6, ...x = pi/12, 5pi/12, 13pi/12, 17pi/12, ...... Show More
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Lavinia Carroll

Feb 20, 2025

sin4x + cos2x = 0 2sin2x . cos2x + cos2x =0 cos 2x ( 2sin 2x +a million) =0 cos 2x =0 2x=nkp + p/2 x =nkp/2 + p/4 2sin 2x +a million =0 sin2x =-a million/2 2x =2kp -p/6 x=kp- p/12 2x =2kp +7p/6 x=kp + 7p/12... Show More
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Lavinia Carroll

Feb 20, 2025

i dont know if this is right but i would make it 2 seperate equations like so:sin4x=0 and -cos2x=0 and if the equation is correct then x should be the same in both meaning u have x... Show More
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