What is the general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 ?

The general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 is

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2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}, β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

Solution steps

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

Solve by substitution

cos (2 β) = \frac{1}{2}, cos (2 β) = - 1

cos (2 β) = \frac{1}{2}, - 36 0^{\circ} \leq β \leq 36 0^{\circ} : β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}

cos (2 β) = - 1, - 36 0^{\circ} \leq β \leq 36 0^{\circ} : β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

Combine all the solutions

β = 3 0^{\circ}, β = 15 0^{\circ}, β = 21 0^{\circ}, β = 33 0^{\circ}, β = - 15 0^{\circ}, β = - 3 0^{\circ}, β = - 33 0^{\circ}, β = - 21 0^{\circ}, β = 9 0^{\circ}, β = 27 0^{\circ}, β = - 9 0^{\circ}, β = - 27 0^{\circ}

cos (2 x) = \frac{3}{7}

0 = 1 - 3 cos (θ)

(tan (\frac{3 a}{2})) tan (\frac{a}{2}) = 3

1 = 3 sin (x)

\frac{sin ( 11 5 ^{\circ} )}{20} = \frac{sin ( B )}{15}

What is the general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 ?
The general solution for 2cos^2(2β)+cos(2β)-1=0,-360<= β<= 360 is