Answer:
All options are correct
Step-by-step explanation:
1. Note that
[tex]\cos 2x=2\cos ^2x-1.[/tex]
2. Substitute previous expression into the equation:
[tex]2\cos ^2x-1+\cos^2x=1,\\ \\3\cos^2x=2,\\ \\\cos^2x=\dfrac{2}{3},\\ \\\cos x=\pm\sqrt{\dfrac{2}{3}}.[/tex]
3. If 0° < x < 360°, then
[tex]x=\arccos \left(\sqrt{\dfrac{2}{3}}\right)\approx 35^{\circ};[/tex];
[tex]x=180^{\circ}-35^{\circ}=145^{\circ};[/tex]
[tex]x=180^{\circ}+35^{\circ}=215^{\circ};[/tex]
[tex]x=360^{\circ}-35^{\circ}=325^{\circ}.[/tex]
All options are true.
Or you can solve this equation graphically. Plot the graph of the function [tex]y=\cos2x+cos^2x[/tex] that represents the left side of the equation and the graph of the function [tex]y=1[/tex] that represents the right side of the equation. their common points are solutions.
