How do you solve $2{{\cos }^{2}}x-5\cos x+2=0?$

Hint: We will convert the given trigonometric equation to a quadratic equation by equating $\cos x$ with a variable $c.$ Then we will solve the quadratic equation for $c$ using the quadratic formula.

Complete step by step solution:
Let us consider the given trigonometric equation $2{{\cos }^{2}}x-5\cos x+2=0.$
We are asked to solve the given trigonometric equation.
Let us first convert the given trigonometric function into a quadratic equation in $c$ by equating $\cos x$ into $c.$
The trigonometric equation will become a quadratic equation $2{{c}^{2}}-5c+2=0.$
We can use the quadratic formula to solve the obtained quadratic equation.
Suppose that we have a quadratic equation $a{{x}^{2}}+bx+c=0.$ Then we can use the following formula which is called the quadratic formula to solve the equation for $x: x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.$
Let us compare the coefficients to get $a=2, b=-5$ and $c=2.$
So, the quadratic formula for our equation is $c=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 2\times 2}}{2\times 2}.$
We can write this equation as \[c=\dfrac{5\pm \sqrt{25-16}}{4}.\]
That can be written as $c=\dfrac{5\pm \sqrt{9}}{4}.$
When we write this equation by substituting $3=\sqrt{9},$ we will get $c=\dfrac{5\pm 3}{4}.$
From this, we will get the possible values of the variable $c$ as $c=\dfrac{5-3}{4}$ or $c=\dfrac{5+3}{4}.$
And this will give us $c=\dfrac{2}{4}=\dfrac{1}{2}$ or $c=\dfrac{8}{4}=2.$
We know that $\cos x=c.$ So, we will get $\cos x=2$ or $\cos x=\dfrac{1}{2}.$
But we know that the range of the Cosine function is $\left[ -1,1 \right].$
So, it is not possible that $\cos x=2.$
Therefore, we will get $\cos x=\dfrac{1}{2}.$
We know that the values of $x$ for which $\cos x=\dfrac{1}{2}$ are $\pm \dfrac{\pi }{3}.$
In general, $x=\pm \dfrac{\pi }{3}+2n\pi ,n\in \mathbb{Z}.$

Hence the solution of the given equation is given by $x=\pm \dfrac{\pi }{3}+2n\pi ,n\in \mathbb{Z}.$

Note: We can factorize the obtained quadratic equation $2{{c}^{2}}-5c+2=0$ as $\left( c-2 \right)\left( 2c-1 \right)=0.$ Then from the first factor, we will get $c=2$ and from the second fraction, we will get $2c=1\Rightarrow c=\dfrac{1}{2}.$