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anonymous
 5 years ago
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)
anonymous
 5 years ago
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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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ugh... thats a nightmare function if a ever seen one

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the last term is +6sin(x)... that parts easy

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.06sqrt(3) is really a constant so its avoidable

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin(x) goes to cos(x); that was easy lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[f'(x) = 6 \sqrt{3} \cos(x)+6\sin(x)=0\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.06(sqrt(3)cos(x) + sin(x)) = 0 sqrt(3)cos(x) = sin(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Should be pi/6, i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, 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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