Question

I don't have a clue please help

solve for x, 0° ≤ x ≤ 360°. Write degree measurements in ascending order.


cos 2 x+cos²x=1 x = ????

Answer 1

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) = (2)/(3)`

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

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

So, the solutions are

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

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

`-35+360 = 325`