Question

Select all that apply. Solve for x , 0 < x < 360;

Answer 1

Answer:

All options are correct

Step-by-step explanation:

1. Note that

$\cos 2x=2\cos ^2x-1.$

2. Substitute previous expression into the equation:

$2\cos ^2x-1+\cos^2x=1,\\ \\3\cos^2x=2,\\ \\\cos^2x=\dfrac{2}{3},\\ \\\cos x=\pm\sqrt{\dfrac{2}{3}}.$

3. If 0° < x < 360°, then

$x=\arccos \left(\sqrt{\dfrac{2}{3}}\right)\approx 35^{\circ};$;

$x=180^{\circ}-35^{\circ}=145^{\circ};$

$x=180^{\circ}+35^{\circ}=215^{\circ};$

$x=360^{\circ}-35^{\circ}=325^{\circ}.$

All options are true.

Or you can solve this equation graphically. Plot the graph of the function $y=\cos2x+cos^2x$ that represents the left side of the equation and the graph of the function $y=1$ that represents the right side of the equation. their common points are solutions.