Answer:
[tex]\large\boxed{x=\pm\dfrac{3\pi}{20}+\dfrac{2n\pi}{5}\ \vee\ x=\pm\dfrac{\pi}{20}+\dfrac{2n\pi}{5}}[/tex]
Step-by-step explanation:
[tex][tex]\cos^2(5x)=\sin^2(5x)\qquad\text{substitute}\ t=5x\\\\\cos^2t=\sin^2t\qquad\text{use}\ \sin^2\theta+\cos^2\theta=1\to\sin^2\theta=1-\cos^2\theta\\\\\cos^2t=1-\cos^2t\qquad\text{add}\ \cos^2t\ \text{to both sides}\\\\2\cos^2t=1\qquad\text{divide both sides by 2}\\\\\cos^2t=\dfrac{1}{2}\Rightarrow \cos t=\pm\sqrt{\dfrac{1}{2}}\\\\\cos t=\pm\dfrac{\sqrt2}{2}\\\\\cos t=-\dfrac{\sqrt2}{2}\Rightarrow t=\pm\dfrac{3\pi}{4}+2n\pi\\\\\cos t=\dfrac{\sqrt2}{2}\Rightarrow t=\pm\dfrac{\pi}{4}+2n\pi[/tex]
[tex]t=5x\\\\5x=\pm\dfrac{3\pi}{4}+2n\pi\ \vee\ 5x=\pm\dfrac{\pi}{4}+2n\pi\qquad\text{divide both sides by 5}\\\\x=\pm\dfrac{3\pi}{20}+\dfrac{2n\pi}{5}\ \vee\ x=\pm\dfrac{\pi}{20}+\dfrac{2n\pi}{5}[/tex]
