What is the general solution for 2cos^2(t)-3cos(t)+1=0 ?

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

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2cos^2(t)-3cos(t)+1=0

2 cos^{2} (t) - 3 cos (t) + 1 = 0

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

Solution steps

2 cos^{2} (t) - 3 cos (t) + 1 = 0

Solve by substitution

cos (t) = 1, cos (t) = \frac{1}{2}

cos (t) = 1 : t = 2 πn

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

Combine all the solutions

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

sin (x + 3 0^{\circ}) = cos (x)

sin (\frac{π}{6} x) = 0

sec (x) - 2 cos (x) = 1

2 sin (x) + 3 = 0

4 cos (2 θ) - 10 cos (θ) + 14 = 7

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