How do you solve $\cos x-3\cos \left( \dfrac{x}{2} \right)=0$ ?

Hint: We know the formula cos 2x is equal to $2{{\cos }^{2}}x-1$ from this formula we can prove that cos x is equal to $2{{\cos }^{2}}\dfrac{x}{2}-1$. Now we can use this formula to solve the given question. We can replace cos x with $2{{\cos }^{2}}\dfrac{x}{2}-1$ , then the equation will convert into a quadratic equation where the variable is $\cos \left( \dfrac{x}{2} \right)$ .

Complete step-by-step answer:
We have to solve $\cos x-3\cos \left( \dfrac{x}{2} \right)=0$
We can write cos x is equal to $2{{\cos }^{2}}\dfrac{x}{2}-1$. Now if we replace cos x with $2{{\cos }^{2}}\dfrac{x}{2}-1$. It will change into a quadratic equation.
$2{{\cos }^{2}}\dfrac{x}{2}-1-3\cos \left( \dfrac{x}{2} \right)=0$
We can see that the above equation is a quadratic equation where the variable is $\cos \left( \dfrac{x}{2} \right)$
The formula roots of any quadratic equation is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where a is the coefficient of ${{x}^{2}}$ , b is coefficient of x and c is the constant in a quadratic equation.
In the equation $2{{\cos }^{2}}\dfrac{x}{2}-1-3\cos \left( \dfrac{x}{2} \right)=0$ , a = 2 , b = -3 and c = -1
Substituting the value of a, b and c in the formula, the roots are
$\dfrac{-3\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 2\times -1}}{2\times 2}=\dfrac{-3\pm \sqrt{17}}{4}$
So the roots are $\dfrac{-3+\sqrt{17}}{4}$ and $\dfrac{-3-\sqrt{17}}{4}$
We know that the range of cos x is from – 1 to 1. So $\dfrac{-3-\sqrt{17}}{4}$ can not be the root of the equation. The only possible root is $\dfrac{-3+\sqrt{17}}{4}$
So the value of $\cos \left( \dfrac{x}{2} \right)$ is equal to $\dfrac{-3+\sqrt{17}}{4}$
So x is equal to $2{{\cos }^{-1}}\dfrac{-3+\sqrt{17}}{4}$ and the general solution is $2n\pi +2{{\cos }^{-1}}\dfrac{-3+\sqrt{17}}{4}$ where n is an integer.

Note: While solving any polynomial equation where the variable is any trigonometric function or exponential function, always check the root whether it lies in the range of that function. For example if the variable is an exponential function, then negative roots will be invalid.