Given the equation:
cos²(x) + 9cos(x) = 5cos(x) + 5
cos²(x) + 9cos(x) - 5 cos(x) - 5 = 0
cos²(x) + 4cos(x) - 5 = 0
[cos(x) + 5][cos(x) - 1] = 0
cos (x) = - 5 x ∉ ℝ; so:
cos (x) = 1
Then;
x = 0, and x = 2π
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