What is the general solution for cos(t)=(sqrt(2))/2 ?

The general solution for cos(t)=(sqrt(2))/2 is t= pi/4+2pin,t=(7pi)/4+2pin

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cos(t)=(sqrt(2))/2

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

t = \frac{π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

Degrees

t = 4 5^{\circ} + 36 0^{\circ} n, t = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

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

General solutions for

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

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

t = \frac{π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

tan^{2} (θ) + 5 tan (θ) + 6 = 0

2 cos^{2} (x) - sin^{2} (x) = 1

2 cos^{2} (θ) - sin (θ) - 1 = 0

sin (t) = - 1

sin (3 x) = 1, 0 \leq x \leq 2 π

What is the general solution for cos(t)=(sqrt(2))/2 ?
The general solution for cos(t)=(sqrt(2))/2 is t= pi/4+2pin,t=(7pi)/4+2pin