Determine the number(s), x, between 0 and 2pi where the line tangent to the given function is horizontal.
f(x) = 5sin(x)-5(square root 3) cos(x)

I know I find f'(x) which is 5(cos(x) + (square root 3) sin(x). Next I am suppose to set the function to 0, however, I am having problem solving for X.

Question

Determine the number(s), x, between 0 and 2pi where the line tangent to the given function is horizontal.
f(x) = 5sin(x)-5(square root 3) cos(x)

I know I find f'(x) which is 5(cos(x) + (square root 3) sin(x). Next I am suppose to set the function to 0, however, I am having problem solving for X.

anonymous · Answer

get rid of the 5 and solve 
$$\cos(x)+\sqrt3\sin(x)=0$$ first

anonymous · Answer

rewrite as one function of sine

anonymous · Answer

Thank you, I got it!

anonymous · Answer

yw