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

tan^{3} (x) = tan (x), 0 \leq x \leq 2 π

sin (x^{2} + 3 x + 5) = 0

sin (θ) (2 sin (θ) + 1) = 0

tan (θ) = 0.437

sin (\frac{x}{3}) = - \frac{3}{2}

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