No. x cannot equal 2 and 7
27: C 28: B 29. A
And I kindly decline the marriage offer lol. If you need an explanation of how I did this then I will gladly do so they are pretty simple.
Hey read the image I have attached. Ignore what I am typing here I need to type 20 characters.
See the picture attached to better understand the problem
we know that
in the right triangle ABC
cos A=AC/AB
cos A=1/3
so
1/3=AC/AB----->AB=3*AC-----> square----> AB²=9*AC²----> equation 1
applying the Pythagoras Theorem
BC²+AC²=AB²-----> 2²+AC²=AB²---> 4+AC²=AB²----> equation 2
substitute equation 1 in equation 2
4+AC²=9*AC²----> 8*AC²=4----> AC²=1/2----> AC=√2/2
so
AB²=9*AC²----> AB²=9*(√2/2)²----> AB=(3√2)/2
the answer isthe hypotenuse is (3√2)/2
15) factor out cos: cos(x)(sin(x)+1)=0
Now this is true when either cos(x)=0 (x=pi/2 and 3pi/2)
Or when sin(x)=-1 (x=3pi/2)
So it's solutions are pi/2 and 3pi/2