Cos квадрат x-sinквадрат x=корень 3 делить на 2.

To solve the equation ${\cos }^{2}x-{\sin }^{2}x=\frac{\sqrt{3}}{2}$ , we will use trigonometric identities to simplify the expression and find the general solution for $x$. 1. Simplify the Equation The left side of the equation, ${\cos }^{2}x-{\sin }^{2}x$, is a standard trigonometric identity for the double angle of cosine: $\cos \left(2x\right)={\cos }^{2}x-{\sin }^{2}x$Substituting this into the original equation, we get: $\cos \left(2x\right)=\frac{\sqrt{3}}{2}$ 2. Solve for the Argument Now, we determine the values of the angle $2x$ that satisfy the equation. On the unit circle, the cosine of an angle equals $\frac{\sqrt{3}}{2}$ at $\pm \frac{\pi }{6}$ (or $\pm {30}^{\circ }$). The general formula for the solution of $\cos \left(\theta \right)=a$ is: $\theta =\pm \arccos \left(a\right)+2\pi k,k\in \mathbb{Z}$Applying this to our equation: $2x=\pm \frac{\pi }{6}+2\pi k$ 3. Isolate $x$ To find $x$, divide the entire equation by $2$: $x=\frac{\pm \frac{\pi }{6}+2\pi k}{2}$ This results in: $x=\pm \frac{\pi }{12}+\pi k,k\in \mathbb{Z}$ Summary of Solutions The solutions can be expressed in radians or degrees:

Would you like me to show the specific steps for finding the values of $x$ within a particular interval, such as $[0,2\pi ]$?