Does anyone know how to solve this 4 cos^2 x=3 where is greater than or equal to 0 and less than or equal to

2 ⤊

twopie

2007-01-09 11:39:51 · 2 answers · asked by TeddyGrahams 1 in Science & Mathematics ➔ Mathematics

4(cos x)^2 = 3
(cos x)^2 = 3/4
cos x = +- sqr(3)/2
x = arccos[+-sqr(3)/2] .

2007-01-09 12:54:39 · answer #1 · answered by tablecloth 1 · 2⤊ 0⤋

I hope that you mean 4(cos x)^2 = 3; if so then
(cos x)^2 = 3/4
cos x = +- sqr(3)/2
x = arccos[+-sqr(3)/2]

2007-01-09 19:58:33 · answer #2 · answered by kellenraid 6 · 0⤊ 0⤋