solve cos(3x)+cos(x)=0

Question

my answer key says the right answer is 90^°+180^°k, 45^°+90^°k but I'm confused about what precisely the k stands for and how they got there

Answer 1

k= any integer,

usually n is used for any integer, but this answer key uses k

cos3x + cosx = 0

cos(2x+x) + cosx = 0

cos2xcosx -sin2xsinx + cosx = 0

[cos^2(x)- sin^2(x)]cosx - 2sinxcosxsinx + cosx = 0

cosx[cos^2(x) -sin^2(x) -2sin^2(x) +1]=0 cosx =0, x=cos^-1(0) = 90 +180n degrees, n=any integer

set the other factor = 0

cos^2(x)-3sin^2(x)=-1

cos^2(x)+ sin^2(x) - 4sin^2(x) =-1

1-4sin^2(x) =-1

2sin^2(x) = 1

sin^2(x) =1/2

sinx = +/- sqr(1/2)

x = sin^-1(sqr.5)= 45 +90n degrees

x = 180n, 45 + 90 degrees

let k=n

x= 90+ 180k, 45 +90k

the answer key is correct, as

cos(3x90) + cos90 = 0+0= 0

cos(3x45) + cos45 =0+0= 0

Answer 2

Write Cos (3x) = 4cos³(x) - 3cos(x)

Now cos(3x) + cos(x) = 4cos³(x) - 2cos(x) = 0

You can write Cos(x) = 0, 2cos²(x) - 1 = 0 implies cos(2x) = 0

Therefore, x = 90^o + 180^ok, 45^o + 90^ok where k is any integer (Z). The same could also be expressed in pi radians by substituting pi/4 = 45^o, pi/2 =90^o and pi = 180^o

Hope this helps.

Answer 3

First of all, I suggest you graph the function f(x)=cos(3x)+cos(x)

You will see that the roots are just exactly as described in our answer key, i.e.

90°+k(any integer)*180° and 45°+k(any integer)*90°

This happens because the trig functions are all periodic and sums of periodic functions exhibit periodicity also. To solve the problem algebraically you need to use the reduction formula for cos(3x)...this is a nuisance but it will work.

Answer 4

The trignometric functions are cyclic - look at the graph of sine or cosine.

The solutions are 90° + multiples (that is the k) of 180°

The solutions are 45° + multiples (that is the k) of 90°