How do you solve \[2{{\sin }^{2}}x=1+\cos x\] for \[{{0}^{\circ }}\le x\le {{180}^{\circ }}\]?

Hint: To solve the given trigonometric equation, we first need to convert \[{{\sin }^{2}}x\] in terms of $\cos x$ using the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Using this, we will obtain a trigonometric equation in terms of $\cos x$. Then we need to substitute $\cos x=t$ to get a quadratic equation in $t$. On solving the quadratic equation, we will get two values of $t$, which will correspond to the values of $\cos x$. Then using the given range, we can determine the final solution of the equation.

Complete step by step answer:
The equation given in the question is
\[2{{\sin }^{2}}x=1+\cos x............(i)\]
As we can observe in the above equation that it contains two trigonometric functions, \[{{\sin }^{2}}x\] and $\cos x$. For solving a trigonometric equation, we first have to convert it in the form of only one trigonometric function. Now, clearly $\cos x$ can’t be converted in any form so it will remain as it is. But we can convert \[{{\sin }^{2}}x\] in terms of $\cos x$ by using the identity
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
Subtracting ${{\cos }^{2}}x$ from both the sides, we get
\[\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x.............(ii)\]
Putting equation (ii) in (i), we get
\[\begin{align}
  & \Rightarrow 2\left( 1-{{\cos }^{2}}x \right)=1+\cos x \\
 & \Rightarrow 2-2{{\cos }^{2}}x=1+\cos x \\
\end{align}\]
Adding $2{{\cos }^{2}}x-2$ both the sides, we get
\[\begin{align}
  & \Rightarrow 2-2{{\cos }^{2}}x+2{{\cos }^{2}}x-2=1+\cos x+2{{\cos }^{2}}x-2 \\
 & \Rightarrow 0=2{{\cos }^{2}}x+\cos x-1 \\
 & \Rightarrow 2{{\cos }^{2}}x+\cos x-1=0..........(iii) \\
\end{align}\]
Substituting $\cos x=t$ in the above equation, we get
\[\Rightarrow 2{{t}^{2}}+t-1=0\]
So we have a quadratic equation in $t$. Now, the above equation can be written as
\[\Rightarrow 2{{t}^{2}}+2t-t-1=0\]
Taking $2t$ common from the first two terms, and $-1$ common from the last two terms, we get
\[\Rightarrow 2t\left( t+1 \right)-1\left( t+1 \right)=0\]
Now taking \[\left( t+1 \right)\] common, we have
\[\begin{align}
  & \Rightarrow \left( t+1 \right)\left( 2t-1 \right)=0 \\
 & \Rightarrow \left( t+1 \right)=0,\left( 2t-1 \right)=0 \\
\end{align}\]
On solving we get
$\Rightarrow t=-1,t=\dfrac{1}{2}$
Now, according to our substitution, $t=\cos x$. This means that
$\cos x=-1,\cos x=\dfrac{1}{2}$
According to the question, the interval of $x$ is given to be \[{{0}^{\circ }}\le x\le {{180}^{\circ }}\]. This means that $x$ can take the values from the first two quadrants.
We know that $\cos x$ is positive in the first quadrant, and is negative in the second quadrant. So the negative value $-1$ must belong to the second quadrant, and the positive value $\dfrac{1}{2}$ must belong to the first quadrant. From the first part of the solution, we have
$\begin{align}
  & \Rightarrow \cos x=-1 \\
 & \Rightarrow x={{180}^{\circ }} \\
\end{align}$
And from the second part of the solution we have
$\begin{align}
  & \Rightarrow \cos x=\dfrac{1}{2} \\
 & \Rightarrow x={{60}^{\circ }} \\
\end{align}$
So the solution of the given equation \[2{{\sin }^{2}}x=1+\cos x\] is $x={{180}^{\circ }}$ and $x={{60}^{\circ }}$.

Note: Do not end your solution after writing the values for $\cos x$.The variable of the equation is $x$, not $\cos x$. So the solution of the equation means the values of the variable $x$. Also, be careful to make sure that the solutions to the given trigonometric equation obtained must be in the range specified in the question.