I know that this question has been answered before but i'm not satisfied with the answer so i'm asking it again. the question is cos4x=cos2x , solve for x Just check the steps
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cos4x-cos2x
= (2cos²(2x) - 1) - cos(2x)
= 2cos²(2x) - cos(2x) - 1
= (2cos(2x) + 1)(cos(2x) - 1)
2cos(2x) = -1
cos(2x) = -1/2
2cos²(x) - 1 = -1/2
2cos²(x) = 1/2
cos²(x) = 1/4
cos(x) = +-1/2
x = π/3, 2π/3, 4π/3, 5π/3
or
cos(2x) = 1
2cos²(x) - 1 = 1
2cos²(x) = 2
cos²(x) = 1
cos(x) = +-1
x = 0, π
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