What is the general solution for 5-20cos^2(t)=0 ?

The general solution for 5-20cos^2(t)=0 is t= pi/3+2pin,t=(5pi)/3+2pin,t=(2pi)/3+2pin,t=(4pi)/3+2pin

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5-20cos^2(t)=0

5 - 20 cos^{2} (t) = 0

t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn, t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Solution steps

5 - 20 cos^{2} (t) = 0

Solve by substitution

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

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

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

Combine all the solutions

t = \frac{π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn, t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

10 tan (B) + 12 = tan (B) + 6

cos (θ) = - \frac{5}{8}

\frac{368.02}{sin ( 6 6 ^{\circ} - x )} = \frac{290}{sin ( 3 8 ^{\circ} )}

cos^{2} (x) sec (x) = 1

tan (x) = \frac{9.3}{7.5}

What is the general solution for 5-20cos^2(t)=0 ?
The general solution for 5-20cos^2(t)=0 is t= pi/3+2pin,t=(5pi)/3+2pin,t=(2pi)/3+2pin,t=(4pi)/3+2pin