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Naddika [18.5K]
3 years ago
14

I don't have a clue please help

Mathematics
1 answer:
Wittaler [7]3 years ago
5 0

First of all, we can use the double angle identity to write

\cos(2x) = 2\cos^2(x)-1

The equation becomes

2\cos^2(x)-1+\cos^2(x) = 1 \iff 3\cos^2(x) = 2

Divide both sides by 3 to get

\cos^2(x) = \dfrac{2}{3}

And finally consider the square root of both terms (don't forget the double sign):

\cos(x) = \pm\sqrt{\dfrac{2}{3}}

So, the solutions are

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

If you want x in [0,360], consider the equivalent angle

-35+360 = 325

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