What is the general solution for sqrt(2)cos(θ)=1,0<= θ<= 360 ?

The general solution for sqrt(2)cos(θ)=1,0<= θ<= 360 is θ=45,θ=315

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sqrt(2)cos(θ)=1,0<= θ<= 360

2 cos (θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 4 5^{\circ}, θ = 31 5^{\circ}

Radians

θ = \frac{π}{4}, θ = \frac{7 π}{4}

Solution steps

2 cos (θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

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

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = 4 5^{\circ}, θ = 31 5^{\circ}

- 2 cos (x) = 1

0 = 6 cos (3 t)

2 cos (x) = sin (x) + 1

cos (z) = \frac{5}{4}

cos (B) = \frac{3}{8}

What is the general solution for sqrt(2)cos(θ)=1,0<= θ<= 360 ?
The general solution for sqrt(2)cos(θ)=1,0<= θ<= 360 is θ=45,θ=315