How do you solve $2{{\cos }^{2}}\theta +\sin \theta =1$? \[\]

Hint: We use the Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to convert the cosine into sine. We replace $\sin \theta =x$ and solve the quadratic equation in $x$ by splitting the middle term method. We equate the roots with $\sin \theta $ and us the standard solutions $\sin \theta =\sin \alpha $ as $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $ where $n$ is any arbitrary integer.\[\]

Complete step-by-step solution:
We know that a trigonometric equation is an equation with trigonometric functions with unknown arguments or angles. When we are asked to solve a trigonometric equation we have to find all possible measures of unknown angles. We are given the following trigonometric equation to solve
$2{{\cos }^{2}}\theta +\sin \theta =1$
We use the Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ to convert the cosine into sine to have;
\[ \begin{align}
  & 2\left( 1-{{\sin }^{2}}\theta \right)+\sin \theta =1 \\
 & \Rightarrow 2-2{{\sin }^{2}}\theta +\sin \theta =1 \\
 & \Rightarrow -2{{\sin }^{2}}\theta +\sin \theta +1=0 \\
 & \Rightarrow 2{{\sin }^{2}}\theta -\sin \theta -1=0 \\
\end{align}\]
We see that the above equation is a quadratic equation in $\sin \theta $. Let us put $t=\sin \theta $ in the above step to have;
\[\Rightarrow 2{{t}^{2}}-t-1=0\]
We shall solve the above quadratic equation by splitting the middle term $-t$into $-2t+t$. We have;
\[\begin{align}
  & \Rightarrow 2{{t}^{2}}-2t+t-1=0 \\
 & \Rightarrow 2t\left( t-1 \right)+1\left( t-1 \right)=0 \\
 & \Rightarrow \left( t-1 \right)\left( 2t+1 \right)=0 \\
 & \Rightarrow t-1=0\text{ or }2t+1=0 \\
\end{align}\]
So we have the two roots of the quadratic equation as
\[\Rightarrow t=1,t=\dfrac{-1}{2}\]
We put back $t=\sin \theta $ in the above step to have
\[\begin{align}
  & \Rightarrow \sin \theta =1,\sin \theta =\dfrac{-1}{2} \\
 & \Rightarrow \sin \theta =\sin \left( \dfrac{\pi }{2} \right),\sin \theta =\sin \left( \dfrac{7\pi }{6} \right) \\
\end{align}\]
We know that the solutions of the equation $\sin \theta =\sin \alpha $ where $\alpha $ is the principal solution are given by $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $. The solutions of the above equation are given by
\[\Rightarrow \theta =n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2} \right),\theta =n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{7\pi }{6} \right)\]
So the general solution set of the given trigonometric equation is
\[\Rightarrow \theta =\left\{ n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2} \right)\bigcup n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{7\pi }{6} \right):n\in \mathsf{\mathbb{Z}} \right\}\]

Note: We note that we can also write $\dfrac{7\pi }{6}=\dfrac{-5\pi }{6}$ by common arc angle and find the solution. We note that the first solution of the trigonometric equation within the interval $\left[ 0,2\pi \right]$is called principal solution and using periodicity all possible solutions obtained with integer $n$ are called general solutions. We should try to convert sine to cosine using $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ because we have to square twice and the solution will be difficult.