Determine the numbers, x, between 0 and 2where the line tangent to the given function is horizontal f(x)=6(sqrt3)sin(x)-6cos(x)

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Determine the numbers, x, between 0 and 2where the line tangent to the given function is horizontal f(x)=6(sqrt3)sin(x)-6cos(x)

Mathematics
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ugh... thats a nightmare function if a ever seen one
the last term is +6sin(x)... that parts easy
6sqrt(3) is really a constant so its avoidable

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Other answers:

sin(x) goes to cos(x); that was easy lol
\[f'(x) = 6 \sqrt{3} \cos(x)+6\sin(x)=0\]
6(sqrt(3)cos(x) + sin(x)) = 0 sqrt(3)cos(x) = sin(-x)
we can square both sides maybe? 3cos^2 = sin^2 3cos^2(x) + sin^2(x) = 0 such odd numbers; you sure it aint 0 to 2pi?
Should be pi/6, i think
ohh it is 2pi
sorry, and the answer choices are a: {2/3pi, 5/3pi} b. {0,1} c. {1/3pi,5/3pi} d.(1/2pi,3/2pi} e. :2/3pi,10/3pi}

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