How do we find angle of like negative integer is there like cosx=-1/3 normally for positive cosx we take angle in first and fourth quadrant but what if it is like I wrote above negative And how to find angle if range start from negative I.e.-180 to 180?

Let's take `cos x = -1/2` first, which is one you're more likely to see.

For triangle angles, between `0` and `180^circ`, the cosine is positive for acute angles, zero for a right angle, and negative for obtuse angles, bigger than `90^circ.` So the negative cosine tells us one solution is in the second quadrant.

Here `-1/2` should tell you it's a 30/60/90 triangle, the biggest cliche in trig. The angle in the second quadrant whose cosine is `-1/2` is

`cos 120^circ = -1/2`

If you didn't know that offhand but remembered

`cos 60^circ = 1/2`

then the rule about cosines of supplementary angles tells us

`cos 120^circ = cos(180^circ - 60^circ) = - cos 60^circ = -1/2`

So now we're faced with

`cos x = cos 120^circ`

In general `cos x = cos a` has solutions `x = pm a + 360^circ k quad` for integer `k`.

Here, we find the general solution to `cos x = -1/2` is

`x = \pm 120^circ + 360^circ k quad` for integer `k`

We can pick out the ones here between `-180^circ` and `180^circ` (which are `-120^circ` and `120^circ`) or between `0` and `360^circ` (which are `120^circ` and `240^circ`).

We see these are in the second and third quadrants, which is where the negative cosines live.

Now let's take the original question,

`cos x = -1/3`

That's not one with a nice form for `x`. We just write the equation using the principal value of the inverse cosine

`cos x = cos ( text{Arc}text{cos}(-1/3))`

and apply our solution

`x = \pm text{Arc}text{cos}(-1/3) + 360^circ k quad` integer `k`

We might also write `x = arccos c` as the general solution to `cos x = c .` In other words, we define

`arccos(a) = pm text{Arc}text{cos}(a) + 360^circ k quad` integer `k`

as a multivalued expression, giving all the solutions to `cos x =cos a`.