How do you solve \[2{{\sin }^{2}}\left( a \right)=2+\cos \left( a \right)\] and find all solutions in the interval \[\left[ 0,2\pi \right)\]?

Hint: In order to find the solution to the given question that is to solve \[2{{\sin }^{2}}\left( a \right)=2+\cos \left( a \right)\] and find all solutions in the interval \[\left[ 0,2\pi \right)\], apply the one of the well-known identities of trigonometry that is \[{{\sin }^{2}}\left( a \right)=1-{{\cos }^{2}}\left( a \right)\]. Substitute this value in the given expression and then find the value of \[\cos \left( a \right)\] followed by the finding the value of \[a\] in the interval \[\left[ 0,2\pi \right)\].

Complete step-by-step solution:
According to the question, given equation in the question is as follows:
\[2{{\sin }^{2}}\left( a \right)=2+\cos \left( a \right)\]
To solve the above equation, apply one of the well-known identities of trigonometry that is \[{{\sin }^{2}}\left( a \right)=1-{{\cos }^{2}}\left( a \right)\] & substitute in the above equation, we get:
\[\Rightarrow 2\left( 1-{{\cos }^{2}}\left( a \right) \right)=2+\cos \left( a \right)\]
Simplify the terms in the left-hand side of the above equation by opening the bracket we get:
\[\Rightarrow 2-2{{\cos }^{2}}\left( a \right)=2+\cos \left( a \right)\]
As we can see term \[2\] can be cancelled from both the sides in the above equation, we get:
\[\Rightarrow -2{{\cos }^{2}}\left( a \right)-\cos \left( a \right)=0\]
Now, take the term \[-\cos \left( a \right)\]in common and rest the terms in the bracket from the above equation, we will have:
\[\Rightarrow -\cos \left( a \right)\left( 2\cos \left( a \right)+1 \right)=0\]
From above equation we can infer two results that are:
1) \[\cos \left( a \right)=0\]
\[\Rightarrow a=\dfrac{\pi }{2}\] and \[a=\dfrac{3\pi }{2}\]
2) \[2\cos \left( a \right)+1=0\]
\[\Rightarrow \cos \left( a \right)=-\dfrac{1}{2}\]
\[\Rightarrow a=\pm \dfrac{2\pi }{3}\]
The arc \[\dfrac{2\pi }{3}\] and arc \[\dfrac{4\pi }{3}\] are coterminal.
Therefore, the values of \[a\] in the interval \[\left[ 0,2\pi \right)\] are \[\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{2\pi }{3}\text{ }\!\!\And\!\!\text{ }\dfrac{4\pi }{3}\].

Note: Students generally make mistakes while finding the value of angle of cosine or any trigonometric function by just finding out one value, but it’s actually two answers. It’s important to check for the 2nd value and don’t just ignore it because it gives precision to the result of your answer and it’s important the remember the identity \[{{\sin }^{2}}\left( a \right)=1-{{\cos }^{2}}\left( a \right)\].