Question

Solve $ \cos 2x + 3\cos x - 1 = 0. $

Answer 1

Hint: By using the basic trigonometric identity given below we can simplify the above expression.
 $ \cos 2x = 2{\cos ^2}x - 1 $
Also Quadratic Formula: $ a{x^2} + bx + c = 0 $ Here $ a,\;b,\;c $ are numerical coefficients.
So to solve $ x $ we have: $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Such that in order to solve and simplify the given expression we have to use the above identity and express our given expression in the quadratic form and thereby solve it.

Complete step-by-step answer:
Given
 $ \cos 2x + 3\cos x - 1 = 0...................\left( i \right) $
Now we have to simplify the given equation and try to express it in the form of a quadratic equation, such that we can solve for $ x $ as shown above.
Now we know:
 $ \cos 2x = 2{\cos ^2}x - 1........................\left( {ii} \right) $
Substituting (ii) in (i) we get:
 $
  \cos 2x + 3\cos x - 1 = 2{\cos ^2}x - 1 + 3\cos x - 1 \\
   = 2{\cos ^2}x + 3\cos x - 2......................\left( {iii} \right) \\
  $
Now we have to express the given expression in quadratic form.
So let
\[
  \,\,\,\,\,\,\,p = \cos x \\
   \Rightarrow 2{\cos ^2}x + 3\cos x - 2 = 2{p^2} + 3p - 2............\left( {iv} \right) \;
 \]
On observing (iv) we can say that it’s in the form of $ a{x^2} + bx + c = 0 $ such that by the quadratic formula we can solve for p.
So we have:
 $ 2{p^2} + 3p - 2\,{\text{where}}\,a = 2,\,b = 3\,{\text{and}}\,c = - 2. $
Now we can solve for $ p $ by using the formula:
 $ p = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}.........................\left( v \right) $
Now substituting the value of ‘a’, ‘b’ and ‘c’ in (v) we get:.
\[
   \Rightarrow p = \dfrac{{ - 3 \pm \sqrt {{3^2} - \left( {4 \times 2 \times - 2} \right)} }}{{2 \times 2}} \\
   \Rightarrow p = \dfrac{{ - 3 \pm \sqrt {9 + \left( {16} \right)} }}{4} \\
   \Rightarrow p = \dfrac{{ - 3 \pm \sqrt {25} }}{4} \\
   \Rightarrow p = \dfrac{{ - 3 \pm 5}}{4} \\
   \Rightarrow p = \dfrac{{ - 3 + 5}}{4}\,{\text{and}}\,\dfrac{{ - 3 - 5}}{4} \\
   \Rightarrow p = \dfrac{2}{4}\,{\text{and}}\, - \dfrac{8}{4} \\
   \Rightarrow p = \dfrac{1}{2}\,{\text{and}}\, - 2...........................\left( {vi} \right) \;
 \]
Now substituting for ’p’ back in (iv) we get:
 $
  p = \cos x = \dfrac{1}{2}..................\left( {vii} \right) \\
  {\text{or}} \\
  p = \cos x = - 2...............\left( {viii} \right) \;
  $
So now on analyzing (vii) and (viii) we see that we now have two possibilities of $ \cos x $ , but we also know that the range of $ \cos x\;{\text{is}}\;\left[ { - 1,1} \right]. $
Such that equation (viii) $ p = \cos x = - 2 $ doesn’t exist and is invalid.
So we have our answer $ p = \cos x = \dfrac{1}{2}. $
Now solving for $ x $ we get:
 $
   \Rightarrow \cos x = \dfrac{1}{2} \\
   \Rightarrow x = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\
   \Rightarrow x = \dfrac{\pi }{3} + 2\pi n,\;\dfrac{{5\pi }}{3} + 2\pi n............................\left( {ix} \right) \;
  $
Therefore our final answer is:
 $ x = \dfrac{\pi }{3} + 2\pi n,\;\;\dfrac{{5\pi }}{3} + 2\pi n $
So, the correct answer is “ $ x = \dfrac{\pi }{3} + 2\pi n,\;\;\dfrac{{5\pi }}{3} + 2\pi n $ ”.

Note: Some other equations needed for solving these types of problem are:
\[
  \begin{array}{*{20}{l}}
  {\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right)} \\
  {\cos \left( {2\theta } \right) = {{\cos }^2}\left( \theta \right)-{{\sin }^2}\left( \theta \right) = 1-2{\text{ }}{{\sin }^2}\left( \theta \right) = 2{\text{ }}{{\cos }^2}\left( \theta \right)-1}
\end{array} \\
    \\
 \]
Range of cosine and sine: $ \left[ { - 1,1} \right] $
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.