What is the general solution for solvefor θ,cos(θ)=(sqrt(2))/2 ?

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

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solvefor θ,cos(θ)=(sqrt(2))/2

solve for

θ, cos (θ) = \frac{2}{2}

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

Degrees

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

Solution steps

cos (θ) = \frac{2}{2}

General solutions for

cos (θ) = \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}

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

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

b, s (t) = tan (b t^{2})

tan (x) = \frac{80}{90}

- \frac{5}{4} = cos (x)

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

4 cos^{2} (x) - 11 cos (x) + 6 = 0

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