Given that #cos(x) = -3/4#; what is the value of #x#?

Question

Simply the question itself, in my book the answer should be #pi/3# but I can't remember how to solve it.
Or maybe the book is wrong in this case...

11415 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-03T01:11:18

#cos(x)=-3/4#

#rArrx=arccos(-3/4)#

#rArrcolor(red)(x~~+-2.418858406)# (in radians)

#-3/4# is not one of the special values of the basic trigonometric functions, so it is difficult to express #x# exactly in this case.

The so-called "special angles" and their corresponding outputs for the cosine function are as follows:

#x=0->cosx=1#

#x=pi/6->cosx=sqrt3/2#

#x=pi/4->cosx=sqrt2/2#

#x=pi/3->cosx=1/2#

#x=pi/2->cosx=0#