Answer:
[tex]\large\boxed{x\approx35^0\ \vee\ x\approx145^o\ \vee\ x\approx215^o\ \vee\ x\approx325^o}[/tex]
Step-by-step explanation:
[tex]\text{From (i) we have}\ \cos^4x=1-2\sin^2x+\sin^4x.\\\\\text{Substitute to the equation}\ 8\sin^4x+\cos^4x=2\cos^2x:\\\\8\sin^4x+1-2\sin^2x+\sin^4x=2\cos^2x\\\\\footnotesize{\text{Use}\ \sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x}\\\\9\sin^4x+1-2\sin^2x=2(1-\sin^2x)\\\\\footnotesize{\text{use the distributive property}\ a(b+c)=ab+ac}\\\\9\sin^4x+1-2\sin^2x=2-2\sin^2x\\\\_{\text{add}\ 2\sin^2x\ \text{from both sides}}\\\\9\sin^4x+1=2[/tex]
[tex]\footnotesize{\text{subtract 1 from both sides}}\\\\9\sin^4x=1\\\\\footnotesize{\text{divide both sides by 9}}\\\\\sin^4x=\dfrac{1}{9}\to\sin x=\pm\sqrt[4]{\dfrac{1}{9}}\\\\\sin x=\pm\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}\\\\\sin x=\pm\dfrac{\sqrt3}{3}\\\\\text{look at the picture}[/tex]
