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

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

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sqrt(2)cos(θ)=1

2 cos (θ) = 1

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

2 cos (θ) = 1

Divide both sides by

2

2 cos (θ) = 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

Simplify

\frac{2 cos ( θ )}{2} = \frac{1}{2}

Simplify

\frac{2 cos ( θ )}{2} : cos (θ)

\frac{2 cos ( θ )}{2}

Cancel the common factor:

2

= cos (θ)

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}

cos (θ) = \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

cos (x) = sin (4 x)

0 = 2 + 4 sin (x)

sin (x) = \frac{1}{5}

cos (x) + cos (x) tan (x) = 0

4 sin^{3} (x) - sin (x) = 0

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