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)

Question

amistre64 · Answer

ugh... thats a nightmare function if a ever seen one

amistre64 · Answer

the last term is +6sin(x)... that parts easy

amistre64 · Answer

6sqrt(3) is really a constant so its avoidable

amistre64 · Answer

sin(x) goes to cos(x); that was easy lol

amistre64 · Answer

$$f'(x) = 6 \sqrt{3} \cos(x)+6\sin(x)=0$$

amistre64 · Answer

6(sqrt(3)cos(x) + sin(x)) = 0

sqrt(3)cos(x) = sin(-x)

amistre64 · Answer

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?

anonymous · Answer

Should be pi/6, i think

anonymous · Answer

ohh it is 2pi

anonymous · Answer

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}