solve the equation for exact solutions in the interval [0, 360)?

Question

2 cos^2x-square root 3 cosx = 0

Answer 1

Replace y=cosx, you have
2y² - sqrt3*y = 0
y(2y-sqrt3) =0

Solutions are y=0 ---> cosx=0 ---> x=90° or x=270°
And y=sqrt3/2 ---> cosx=sqrt3/2 ---> x=30° or x=330°

Answer 2

I assume the notation means 2cos²x - root3 cosx = 0.

We can factorize out cosx from all the terms (don't divide by cosx! You will lose a solution) to give

cosx(2cosx - root3) = 0

Therefore either cosx = 0 or 2cosx - root3 = 0

Therefore either cosx = 0 or cosx = (root3) / 2

If cosx = 0, then x = 90º.

If cosx = (root3) / 2, then x = 30º or 360 - 30º (this can be shown by the definition of cosine)

Therefore x = 30º, 90º, or 330º.

Whoops, yes, sorry, of course cosx = 0 gives x = 270º as well *blush* Well done benoit3535

Answer 3

2 cos^2x-square root 3 cosx = 0
cosx(2cos x-sqrt(3) =0
cos x=0 , so x 90 or 270 degrees
2cosx - sqrt(3) =0
2cos x = sqrt(3)
cos x=sqrt(3)/2, so x = 30 and 330 degrees

Answer 4

y=0
x=90, 330, and 270 degrees.

Answer 5

you mean 2cos^(2x)(x) - squareroot of 3cos(x) = 0 , in the interval [0, 2Pi)?