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

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

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

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

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

Degrees

t = 13 5^{\circ} + 36 0^{\circ} n, t = 22 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{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn

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

2 sin^{2} (3 x) = 1

sin^{2} (θ) - 16 = 0

sin (x) + tan (x) = 0

sin^{2} (x) cos^{2} (x) = 0

3 sin (x) cos (x) + 5 cos (x) = 0

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