Explanation
Given the trigonometric expression
[tex]2cos^2\theta-2sin^2\theta=\sqrt{2}[/tex]
We can find the solutions below;
From trigonometry identities;
[tex]cos^2\theta-sin^2\theta=cos2\theta[/tex]
Therefore; we will have;
[tex]\begin{gathered} 2cos^2\theta-2s\imaginaryI n^2\theta=\sqrt{2} \\ 2(cos^2\theta-sin^2\theta)=\sqrt{2} \\ 2cos2\theta=\sqrt{2} \\ Divide\text{ both sides by 2} \\ \frac{\begin{equation*}2cos2\theta\end{equation*}}{2}=\frac{\sqrt{2}}{2} \\ cos2\theta=\frac{\sqrt{2}}{2} \end{gathered}[/tex]
Therefore, the general solutions for the above is given as;
[tex]\begin{gathered} 2θ=\frac{\pi}{4}+2\pi n,\:2θ=\frac{7\pi}{4}+2\pi n \\ therefore; \\ \theta=\frac{\frac{\pi}{4}+2\pi n}{2},\theta=\frac{\frac{7\pi}{4}+2\pi n}{2} \\ \theta=\frac{\pi}{8}=\pi n,\theta=\frac{7\pi}{8}+\pi n \end{gathered}[/tex]
Answer:
[tex]\theta=\frac{\pi}{8}+\pi n,\theta=\frac{7\pi}{8}+\pi n[/tex]
