The fisrt thing to note is that there is no variable in the question and so there cannot be a solution set. The only possibilities are the statement is true, false or indeterminate.
Now, 2 cos 0 - 1 = 0
is equivalent to 2*1 - 1 = 0 [since cos(0) = 1]
or 2 - 1 = 0
which is false.
cos x - 0.5 = 0 ⇒ cos x = 0.5 ⇒ x = 2nπ ± π/3
either cos OR tan-sin equals zero socos=0 at pi/2 and 3pi/2ortan=sin which is impossibleim not sure though
(0, -14)
x = 0 or x = 2
(-2,-1)
cos x - 0.5 = 0 ⇒ cos x = 0.5 ⇒ x = 2nπ ± π/3
either cos OR tan-sin equals zero socos=0 at pi/2 and 3pi/2ortan=sin which is impossibleim not sure though
a = [ 0, 4 ]
x-2y=0 x=2y The solution set is the set of all (x,y) such that x=2y
(0, -14)
x = 0 or x = 2
'x' must be zero, so the solution set is [ 0 ].
(-2,-1)
17
Using the quadratic formula, I found the solution set is x=2,x=-9
Not sure about the set builder notation, but Q = {0}, the set consisting only of the number 0.
cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...