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?

1 Answer
May 8, 2018

# cos x = -1/3# has solutions

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

Explanation:

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#.