I need to solve this:

arccos4x-2=0

@amistre64 could you please have a look?

Question

I need to solve this:

arccos4x-2=0

@amistre64 could you please have a look?

anonymous · Answer

dude it is so simple place the 2 at the other hand

anonymous · Answer

and then solve for it

amistre64 · Answer

is the -2 part of the arguement of the arccos?

kamille · Answer

no, but actually I made the problem incorrectly. It actually asks:

When the equation has any solutions? (I need to find 'meanings' of a, if I understood correctly now)

$$acos4x-2=0$$

kamille · Answer

I know that Ef of cosx must be [-1;1]

kamille · Answer

and only then it could have any solutions?

amistre64 · Answer

not sure i understand the nature of the question yet.  are you able to post it as a picture? to omit lost in translation errors?

kamille · Answer

$$-1\le a*4x-2<1$$

?

kamille · Answer

ah,but the question is in my own language, you can check, it is 4b http://i.imgur.com/ZQDo1uz.png

kamille · Answer

it says 'which meanings 'a' should have so equation has any solutions'

amistre64 · Answer

google says its luthuainian

kamille · Answer

haha yes

kamille · Answer

sorry! anyways, thanks,I will try to do myself, I know the answer

amistre64 · Answer

a cos(4x)-2 = 0

a cos(4x) = 2

a = 2 sec(4x)

amistre64 · Answer

its just a simple algebra process

kamille · Answer

but the answer is $$a \in (-\infty;-2]U[2;+\infty)$$

amistre64 · Answer

sec(4x) has some values of x to omit, namely some modified version of odd multiples of pi/2 attributed to the period adjustment

amistre64 · Answer

sec(4x) ~~ cos(4x) = 0 when 

4x = (2n+1)pi/2

x = (2n+1)pi/8

amistre64 · Answer

so the range of  2 sec(4x) is the  solution to a

amistre64 · Answer

regardless of the period,  the range is +- 2 to +- inf

amistre64 · Answer

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