Question

How do you solve $\cos x - \cos 2x = 0?$

Answer 1

Hint : As we know that the above given equation contains trigonometric functions. We need to solve so we will use the trigonometric identities to get the solution. Trigonometric equations that have multiple angle terms like in the above equation can be simplified using trigonometric identities. We have to isolate a single trigonometric function on one side of the equation to get the solution.

Complete step-by-step answer :
Here we have an equation $\cos x - \cos 2x = 0$ . We know the identity that,
$\cos 2x = 2{\cos ^2}x - 1$ . So we will apply this in the given equation and put the value of
$\cos 2x$ we get: $\cos x - (2{\cos ^2}x - 1) = 0$ , by breaking the bracket we get,
$\cos x - 2{\cos ^2} + 1 = 0$ . We will interchange the sign and give it the form of the quadratic equation.
On further solving we have a quadratic equation and then factorise it by splitting the middle term i.e.
$- \cos x = - 2\cos x + \cos x$ and $- 2{\cos ^2}x = - 2\cos x \times \cos x$ .
So we can write the equation as
$2{\cos ^2}x - 2\cos x + \cos x - 1 = 0$ . Therefore by taking the common factors out we have:
$2{\cos ^2}x - \cos x - 1 = 0 \Rightarrow (2\cos x + 1)(\cos x - 1) = 0$ , We have two factors now, solving each we get:
$\cos x - 1 = 0 \Rightarrow \cos x = 1$ , So $x = 0$
And another factor we have
$2\cos x + 1 = 0 \Rightarrow 2\cos x = - 1$ . Therefore
$\cos x = - \dfrac{1}{2}$ , Value of $x = \\{ \dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}\\}$ .
Since there is no restriction given for the domain, any multiple of $2\pi$ can be added to the solution to find another solution, hence the full solution will be like this,
$x = 0 + 2\pi n,n = 0, \pm 1, \pm 2,...$ and then second set will be like
$x = \dfrac{{2\pi }}{3} + 2\pi n,n = 0, \pm 1, \pm 2,..$ and then we have another solution
$x = \dfrac{{4\pi }}{3} + 2\pi n,n = 0, \pm 1, \pm 2,...$
Hence the answer is $\\{ \pm \dfrac{{2\pi }}{3} + 2\pi n,0 + 2\pi n\\}$ .
So, the correct answer is “ $\\{ \pm \dfrac{{2\pi }}{3} + 2\pi n,0 + 2\pi n\\}$ ”.

Note : We should note that we have to start taking values from $n = 0$ . WE can also further check the answer by putting the value of $n$ if $\cos x - \cos 2x = 0$ . We should never forget to take both the negative and positive signs in the general form of the cosine function. We should always remember the general solutions of trigonometric functions before solving the sums as the general solution of cosine function at $0$ is $(n\pi + \dfrac{\pi }{2})$ , also when cosine function is $1$ we have $x = 2n\pi$ .