Answer:
Step-by-step explanation:
Given the expression cos(2β)=√3/2 for 0≤β<2π, we are to find the value of β within the range that satisfies the equation.
[tex]cos(2\beta)=\sqrt{ 3}/2\\\\take \ the\ arccos\ of \ both \ sides\\\\cos^{-1}cos(2\beta) = cos^{-1}\sqrt{{3} }/2 \\ \\2\beta = cos^{-1}\sqrt{{3} }/2 \\\\2\beta = 30^0\\\\\beta = 30/2\\\\\beta = 15^0[/tex]
Since cos id positive in the 4th quadrant, [tex]\beta = 360^0-15^0[/tex], [tex]\beta = 345^0[/tex]
Hence the value of [tex]\beta[/tex] that satisfy the equation are 15° and 345°
Converting to radians;
180° = πrad
15° = 15π/180 rad
15° = π/12 rad
345° = 345π/180
345° = 23π/12 rad
The values in radians are π/12 rad and 23π/12 rad
