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=(2pi)/3+2pin,t=(4pi)/3+2pin,t=pi+2pin

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

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

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn, t = π + 2 πn

Solution steps

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

Solve by substitution

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

cos (t) = - \frac{1}{2} : t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

cos (t) = - 1 : t = π + 2 πn

Combine all the solutions

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn, t = π + 2 πn

cos (x) = - \frac{1}{5}

tan (\frac{x}{2}) = sin (x)

7 sin (B) + 23 = 0

cos^{2} (3 x) = 1

tan (\frac{π}{4} - x) + cot (\frac{π}{4} - x) = 4

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=(2pi)/3+2pin,t=(4pi)/3+2pin,t=pi+2pin