How do you solve $2{{\cos }^{2}}x-2{{\sin }^{2}}x=1$ and find all solutions in the interval $0\le x<360$ ?

Hint: We have been given an inequation consisting of two trigonometric functions, sine of x and cosine of x. Since the mentioned interval is $0\le x<360$, therefore, we shall only find the principal values only for which this equation holds true. Firstly, we shall take 2 common terms on the left hand side of the equation and then we shall use trigonometric properties to find the final solution.

Complete step by step solution:
Given that $2{{\cos }^{2}}x-2{{\sin }^{2}}x=1$.
We shall take 2 common from the terms in the left hand side of the equation.
$\Rightarrow 2\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)=1$
Dividing both sides by 2, we get
$\Rightarrow {{\cos }^{2}}x-{{\sin }^{2}}x=\dfrac{1}{2}$
Now, we shall use a trigonometric property where the square of cosine of x minus the square of sine of x equal to cosine of twice of x, that is, ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$. Substituting the value of ${{\cos }^{2}}x-{{\sin }^{2}}x$ from this identity, we get
$\Rightarrow \cos 2x=\dfrac{1}{2}$
We know that $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ and $\cos \dfrac{5\pi }{3}=\dfrac{1}{2}$
Therefore, $2x=\dfrac{\pi }{3}$ and $2x=\dfrac{5\pi }{3}$
Dividing 2 from both sides of this equation, we get
 $\Rightarrow x=\dfrac{\pi }{6}$ and $x=\dfrac{5\pi }{6}$
Thus, $x=\dfrac{\pi }{6},\dfrac{5\pi }{6}$.
Therefore, the solution of $2{{\cos }^{2}}x-2{{\sin }^{2}}x=1$ in the interval $0\le x<360$ is $x=\dfrac{\pi }{6},\dfrac{5\pi }{6}$.

Note: In order to find the solution of various trigonometric equations, we must have prior knowledge of the main trigonometric identities. Also, we could have used another trigonometric property, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to substitute the square of sine of x with 1 minus the square of cosine of x, that is, ${{\sin }^{2}}x=1-{{\cos }^{2}}x$. Using this a quadratic equation would have formed in terms of sine of x which could be solved further to get the final solution.