4(cosx)^2 - 3 = 0
Add 3 to each side
4(cosx)^2 = 3
Divide both sides by 4
(cosx)^2 = 3/4
Take the square root of each side
cosx = sqrt(3/4) or -sqrt(3/4)
cosx = sqrt(3)/2 or -sqrt(3)/2
x = pi/6, 11pi/6 or 5pi/6, 7pi/6
:)
Solve for exact or to 1d.p of x E [0 ,2pi]
4cos^2x-3
2 answers
To "solve" we need an equation.
I will assume you meant:
4cos^2 x - 3 = 0
cos^2 x = 3/4
cosx = ± √3/2 , ----> all 4 quadrants
you should recognize that cos π/6 = √3/2
x = π/6
or
x = π - π/6 = 5π/6
or
x = π + π/6 = 7π/6
or
x = 2π - π/6 = 11π/6
I will assume you meant:
4cos^2 x - 3 = 0
cos^2 x = 3/4
cosx = ± √3/2 , ----> all 4 quadrants
you should recognize that cos π/6 = √3/2
x = π/6
or
x = π - π/6 = 5π/6
or
x = π + π/6 = 7π/6
or
x = 2π - π/6 = 11π/6
4 answers