What is the general solution for 2cos^2(a)+cos(a)-1=0 ?

The general solution for 2cos^2(a)+cos(a)-1=0 is a= pi/3+2pin,a=(5pi)/3+2pin,a=pi+2pin

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2cos^2(a)+cos(a)-1=0

2 cos^{2} (a) + cos (a) - 1 = 0

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn, a = π + 2 πn

Solution steps

2 cos^{2} (a) + cos (a) - 1 = 0

Solve by substitution

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

cos (a) = \frac{1}{2} : a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

cos (a) = - 1 : a = π + 2 πn

Combine all the solutions

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn, a = π + 2 πn

6 sin (t) = 3

arctan (e^{x}) = \frac{π}{4}

csc^{2} (x) + 2 cos (x) + 1 = 0

2 cos^{2} (2 x) + 3 cos (2 x) - 2 = 0

tan (t) = - \frac{12}{5}, \frac{3 π}{2} < t < 2 π

What is the general solution for 2cos^2(a)+cos(a)-1=0 ?
The general solution for 2cos^2(a)+cos(a)-1=0 is a= pi/3+2pin,a=(5pi)/3+2pin,a=pi+2pin