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Determine the numbers x between 0 ad 2pi where the line tangent to the curve is horizontal. y=cos(x)-sqrt(3)sin(x)
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I'm not sure how to approach this one. I've found the derivative y wrt x, which is (-sin(x)-sqrt(3)cos(x). Where would I go from there?
Plug in dy/dx = 0 and solve (because gradient of tangent = 0)
So, set it up like (-sinx-sqrt(3)cosx=0?
Yeah -sin x - √3 cos x = 0 sin x + √3 cos x = 0 R sin (x + a) = 0 R = √(1 + 3) = 2, a = arctan(√3) = π/3 2 sin (x + π/3) = 0 sin (x + π/3) = 0 x + π/3 = 0, π, 2π x = -π/3 (NA) , 2π/3, 5π/3
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