Answer

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**Hint:**At first, we apply the formula $\cos 2x=2{{\cos }^{2}}x-1$ . We get a quadratic in $\cos x$ . We then apply the Sridhar Acharya formula to get the roots of $\cos x$ in the interval $\left[ 0,2\pi \right]$ .

**Complete step by step solution:**

The given equation is

$\cos 2x=2\cos x-1$

If we carefully observe the above equation, we can see that if we can express $\cos 2x$ in terms of $\cos x$ , then the equation becomes an equation of $\cos x$ . Luckily, we have a formula between $\cos 2x$ and $\cos x$ which is,

$\cos 2x=2{{\cos }^{2}}x-1$

Putting this value of $\cos 2x$ in the given equation, the equation thus becomes,

$\Rightarrow 2{{\cos }^{2}}x-1=2\cos x-1$

Now, we subtract $2\cos x$ from both sides of the above equation and get,

$\Rightarrow 2{{\cos }^{2}}x-2\cos x-1=-1$

Now, we cancel $1$ from both sides of the above equation and get,

$\Rightarrow 2{{\cos }^{2}}x-2\cos x=0$

We now divide the entire equation by $2$ and get,

$\Rightarrow {{\cos }^{2}}x-\cos x=0$

This is nothing but a quadratic equation in $\cos x$ . Let us take $\cos x=z$ . We then rewrite the entire equation as,

$\Rightarrow {{z}^{2}}-z=0$

Thus, the quadratic equation has transformed into a quadratic equation in $z$ . So, we now need to solve for $z$ by solving this quadratic. We apply the Sridhar Acharya formula which is

$z=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

In our problem, $a=1,b=-1,c=0$ . So, putting these values in the formula, we get,

$\begin{align}

& \Rightarrow z=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 0 \right)}}{2} \\

& \Rightarrow z=\dfrac{1\pm 1}{2} \\

& \Rightarrow z=\dfrac{1}{2}\pm \dfrac{1}{2} \\

\end{align}$

So, we get two roots of $z$ which are $\dfrac{1}{2}+\dfrac{1}{2}=1$ and $\dfrac{1}{2}-\dfrac{1}{2}=0$ . But, $z=\cos x$ . This means, $\cos x$ has two roots,

$\begin{align}

& \cos x=1...\left( 1 \right) \\

& \cos x=0...\left( 2 \right) \\

\end{align}$

$\left( 1 \right)$ gives $x={{\cos }^{-1}}1$ which gives the values $0,2\pi $ within $\left[ 0,2\pi \right]$ .

$\left( 2 \right)$ gives $x={{\cos }^{-1}}0$ which gives the values $\dfrac{\pi }{2},\dfrac{3\pi }{2}$ within $\left[ 0,2\pi \right]$ .

**Therefore, we can conclude that the values of $x$ in $\left[ 0,2\pi \right]$ which satisfy the given equation are $0,\dfrac{\pi }{2},\dfrac{3\pi }{2},2\pi $.**

**Note:**We should be careful while solving the quadratic equation in $\cos x$ and should apply the Sridhar Acharya formula correctly. This equation can also be solved by taking two equations $y=\cos 2x$ and $y=2\cos x-1$ . The points where two curves intersect in $\left[ 0,2\pi \right]$ will be the required answers.

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