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Sloan [31]
3 years ago
7

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

Mathematics
1 answer:
Vesna [10]3 years ago
8 0

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.

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