Solve the eqution
cos4x + cos2x - 2 =0

cos4x = cos2x

We know that:

cos2x = 1-2cos^2 x

==> cos4x = 1-2cos^2 (2x)

Now substitute:

==> 1-2cos^2 (2x) = cos2x

==> 2cos^2 (2x) + cos2x - 1 = 0

Now factor:

==> (2cos2x -1)(cos2x + 1) = 0

==> 2cos2x -1 = 0 ==> cos2x =1/2 ==> 2x= pi/3

==> x1= pi/6 , 7pi/6

==> x1= pi/6 + 2npi

==> x2= 7pi/6 + 2npi

==> cos2x = -1 ==> 2x= pi ==> x3 = pi/2 + 2npi.

==> x= { pi/6+2npi, 7pi/6+2npi, pi/2+2npi}

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

you can't solve x because it is 2 linear equation and you only have one equation