Question

please show me how to do this problem!? solve for x

Answer 1

\[\cos ^{2}(x)=\cos(x)+2\]

Answer 2

Subtract cos^2(x) from both sides
\[\ 0 = -cos^2(x) + cos(x) + 2\]Factor the problem as you would factor a quadratic. I recommend substituting a value (say "y") for cos(x).
\[y=\cos(x)\]\[-y^2 + y + 2 = 0\]\[(-y+2)(y+1) = 0\]Substitute cos(x) back in for y
\[(-(\cos(x)+2))(\cos(x)+1) = 0\]Since the maximum value for cos(x) is 1 you can throw the first solution out
\[\cos(x) + 1 = 0\]\[\cos(x) = -1\]Solve for x and you get (pi/2) radians or 90 degrees (unit circle is key!)

Answer 3

NOTE: I accidentally wrote "(-(cos(x) +2))" at one point. This is incorrect.
The notation should be "(-cos(x) + 2)" so that the negative sign doesn't distribute to the 2.
If you're still confused just let me know.

Answer 4

the part that i need is when letting y=cosx, thank you, i completely forgot that i could do that