Question

How do you solve $\cos 2x + 5\cos x + 3 = 0$?

Answer 1

In this question, we are given a trigonometric equation and we have been asked to solve it. Start by putting a formula of $\cos 2x$, such that $\cos x$ appear in squares. Then, substitute $\cos x = n$. You will get a quadratic equation. Solve it using quadratic formula. You will get two values of $n$. Put them equal to $\cos x$. Find the value of $x$ using this.

Formula used: $\cos 2x = 2{\cos ^2}x - 1$

Complete step-by-step solution:
We are given a trigonometric equation.
$\Rightarrow \cos 2x + 5\cos x + 3 = 0$
Now, I cannot see any way to start with this equation. Only way I can see is to use the formula of $\cos 2x$.
Putting $\cos 2x = 2{\cos ^2}x - 1$ in the above equation,
$\Rightarrow 2{\cos ^2}x - 1 + 5\cos x + 3 = 0$
Rearranging the terms, we get,
$\Rightarrow 2{\cos ^2}x + 5\cos x + 3 - 1 = 0$
Let us subtract the term and we get
$\Rightarrow 2{\cos ^2}x + 5\cos x + 2 = 0$
Now, we will put $\cos x = n$.
$\Rightarrow 2{n^2} + 5n + 2 = 0$
Now, we have got a quadratic equation. We will solve it using quadratic formula.
The standard equation of quadratic equation is $a{x^2} + bx + c = 0$. On comparing both the equations, we will get –
$\Rightarrow a = 2$, $b = 5$ and $c = 2$.
Now, we will simply put the values in the formula.
$\Rightarrow n = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Putting the values,
$\Rightarrow n = \dfrac{{ - 5 \pm \sqrt {{5^2} - 4 \times 2 \times 2} }}{{2 \times 2}}$
On simplifying the equation, we get,
$\Rightarrow n = \dfrac{{ - 5 \pm \sqrt {25 - 16} }}{4}$
$\Rightarrow n = \dfrac{{ - 5 \pm \sqrt 9 }}{4}$
We can write $\sqrt 9$ as $3$.
Putting in the formula,
$\Rightarrow n = \dfrac{{ - 5}}{4} \pm \dfrac{3}{4}$
Simplifying further,
$\Rightarrow n = \dfrac{{ - 8}}{4},\dfrac{{ - 2}}{4}$
$\Rightarrow n = - 2,\dfrac{{ - 1}}{2}$
Hence, our two values are $- 2, \dfrac{{ - 1}}{2}$.
But, these are the values of $n$. We have to find the values of $x$,
We know that $\cos x = n = - 2,\dfrac{{ - 1}}{2}$ .
Now, let us look at each value individually.
$\Rightarrow \cos x = - 2$ (we will have to reject this value as $\cos x$ cannot be less than $- 1$)
$\Rightarrow \cos x = \dfrac{{ - 1}}{2}$
$\Rightarrow \cos x = \cos \left( {\pi - \dfrac{\pi }{3}} \right)$
$\Rightarrow \cos x = \cos \dfrac{{2\pi }}{3}$
Now, for this value, we are not given any range for which we have to choose the value of x. So, we will give a general value.
If $\cos x = \cos \alpha$, then $x = 2n\pi \pm \alpha$.
Hence, $\cos x = \cos \dfrac{{2\pi }}{3}$

Therefore, $x = 2n\pi \pm \dfrac{{2\pi }}{3}$.

Note: 1) How to know that we have to put the value of $\cos 2x$? While beginning with the question, we see all the possible approaches of the question. The only approach I could see was to expand the term $\cos 2x$ as there was no other way out. So, find all the approaches, look what might be the result of using that particular approach and use the most appropriate solution.
2) Since we were not given any range within which our x lies, we had to find a general solution. If we had given a particular range, then we would have found the values within that range only.