Alan

See the answers to the same question here: 

and here:  

Unfortunately, no!

Choose values of angle θ to go from 0 to 360° and then use x = r*cos(θ) and y = r*sin(θ) where r is the radius of your circle. That's assuming the centre of your circle is at (0,0). If the circle is centred at (xc, yc) you will need to add xc to the x values and yc to the y values.

There are an infinite number of real solutions to this (to say nothing of the complex solutions) as can be seen from the graph below.  If you are looking for real solutions in the range 0 to 2pi (i.e. 0 to 360°)  the graph should help.

P and NP refer to types of solutions to numerical algorithms.  If an algorithm can be solved in "polynomial time" it is referred to as P, if not, it is "non-polynomial time", or NP.  Do an internet search for more detail!

The two are not the same. 

Ermm!  The equation says the angle whose tangent is x is 7/11.

so x = tan(7/11)   (presumably the angle is in radians.)

Or did the questioner mean tan(x) = 7/11, in which case Chris's answer is correct. 

Remember that cosine must have values between ±1 (inclusive).  What is the value of cos(2x) when x is zero, and what is the value of cos(2x) when x is pi/2?

Here are two pictures that might help. The first should enable you to get the magnitude of the exact value, and the second should enable you to get the sign right (remember that an obtuse angle is between 90 and 180).

Since you are given angle A, I suspect you mean angles B and C!

Let 'a' be the length of the side from B to C.  The law of cosines says:

a² = b² + c² - 2*b*c*cos(A)

Put the known information in here:

a² = 6² + 7² -2*6*7*cos(40°)

so 'a' is the square root of the right-hand side:

$${\mathtt{a}} = {\sqrt{{{\mathtt{6}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{7}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{7}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{40}}^\circ\right)}}} = {\mathtt{a}} = {\mathtt{4.544\: \!476\: \!513\: \!087\: \!508\: \!5}}$$

Now we can use the cosine rule in a rearranged form to get cos(B):

cos(B) = (a² + c² - b²)/(2*a*c) or B = cos^-1((a² + c² - b²)/(2*a*c))

$${\mathtt{B}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{\left({{\mathtt{4.544\: \!476\: \!513}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{7}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{6}}}^{{\mathtt{2}}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{4.544\: \!476\: \!513}}{\mathtt{\,\times\,}}{\mathtt{7}}\right)}}\right)} = {\mathtt{B}} = {\mathtt{58.066\: \!464\: \!967\: \!59^{\circ}}}$$

You could use the cosine rule again changing 'b' for 'c' to find angle C if you like.  However, it's probably easier to get it by knowing that A + B + C = 180°

$${\mathtt{C}} = {\mathtt{180}}{\mathtt{\,-\,}}{\mathtt{40}}{\mathtt{\,-\,}}{\mathtt{58.066\: \!465}} = {\mathtt{C}} = {\mathtt{81.933\: \!535}}$$ °

Also, having obtained the size of 'a' it might have been easier to use the sine rule to find B (sin(B)/b = sin(A)/a).