Question

How do you solve $ {\cos ^2}x + 2\cos x + 1 = 0 $ over the interval $ 0 $ to $ 2\pi $ ?

Answer 1

Hint: Try to look at this equation as a quadratic equation and then find its solutions with the conventional method.
A quadratic equation is any equation which can be derived in the form of $ a{x^2} + bx + c = 0 $
Where, $ a,b,c = $ known numbers,
      $ a \ne 0 $
And, $ x = $ unknown or what we call as a “variable”
Solution of a quadratic equation- $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $

Complete step by step answer:
(i)Given equation is,
 $ {\cos ^2}x + 2\cos x + 1 = 0 $
Let $ \cos x $ be $ u $ for the mean time to see the quadratic equation clearly. Thus, substituting $ u $ as a variable in place of $ \cos x $ we get,
 $ {u^2} + 2u + 1 = 0 $
(ii)
As we can clearly see our quadratic equation now, we’ll move forward to find its solutions.
Since, the solutions of a quadratic equation $ a{x^2} + bx + c = 0 $ are:
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
We will substitute the respective values of $ a,b,c $ from our equation $ {u^2} + 2u + 1 = 0 $
So, we have:
  $
  a = 1 \\
  b = 2 \\
  c = 1 \\
  $
Substituting values of $ a,b,c $ in the solution, we get:
 $ u = \dfrac{{ - 2 \pm \sqrt {{2^2} - (4 \times 1 \times 1)} }}{{(2 \times 1)}} $
Solving further,
\[
  u = \dfrac{{ - 2 \pm \sqrt {4 - 4} }}{2} \\
  u = \dfrac{{ - 2 \pm \sqrt 0 }}{2} \\
  u = \dfrac{{ - 2 \pm 0}}{2} \\
  u = \dfrac{{ - 2}}{2} \\
  u = - 1 \\
 \]
Since we got only one value of $ u $ through the formula, we say that this quadratic equation has two equal roots $ - 1, - 1 $
(iii)
Since we assumed $ \cos x $ as $ u $ , we get:
 $ \cos x = - 1 $ i.e., we have to find such value of $ x $ for which $ \cos x $ becomes $ - 1 $ in the interval $ (0,2\pi ) $
Now,
As we know that $ \cos 0 = 1 $
And we also know the identity $ \cos (\pi + \theta ) = - \cos \theta $
Let $ \theta = 0 $
We get, $ \cos (\pi + 0) = - \cos 0 $
i.e., $ \cos \pi = - 1 $
Therefore, for $ x = \pi $ , value of $ \cos x $ becomes $ - 1 $ and $ x = \pi $ clearly lies between $ (0,2\pi ) $ .
Hence, $ x = \pi $ is the solution for $ {\cos ^2}x + 2\cos x + 1 = 0 $.

Note:
The equation given in the question can also be seen as the expanded form of $ {(\cos x + 1)^2} $ . Thus, we can also directly factorize it and equate it with $ 0 $ obtaining the equation $ \cos x = - 1 $ at the end without applying $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ for the quadratic equation. Also, do keep the given range or interval of the solution in mind and make sure you obtain all the solutions lying in that interval only.