What is the general solution for 1/(sqrt(2))=cos(θ) ?

The general solution for 1/(sqrt(2))=cos(θ) is θ= pi/4+2pin,θ=(7pi)/4+2pin

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1/(sqrt(2))=cos(θ)

\frac{1}{2} = cos (θ)

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

\frac{1}{2} = cos (θ)

Switch sides

cos (θ) = \frac{1}{2}

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

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

tan^{2} (x) + 5 tan (x) = 0

1 - sinh^{2} (x) = 0

sin (y + 1 0^{\circ}) = cos (y + 2 0^{\circ})

tan (x) = - 1, 7

tan (θ) (3 cos^{2} (θ) - sin^{2} (θ)) = 0

What is the general solution for 1/(sqrt(2))=cos(θ) ?
The general solution for 1/(sqrt(2))=cos(θ) is θ= pi/4+2pin,θ=(7pi)/4+2pin