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

The general solution for cos(θ)=(sqrt(2))/2 ,sin(θ)=-(sqrt(2))/2 is No Solution for θ\in\mathbb{R}

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cos(θ)=(sqrt(2))/2 ,sin(θ)=-(sqrt(2))/2

cos (θ) = \frac{2}{2}, sin (θ) = - \frac{2}{2}

No Solution for θ \in R

Solution steps

cos (θ) = \frac{2}{2}, sin (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

Solutions for the range

sin (θ) = - \frac{2}{2}

No Solution for θ \in R

cos (θ) = \frac{198.36}{229.4}

(cos (x) - 1) (tan (x) - 1) = 0

- cos (2 x + 3) = 0

tan (2 x) = \frac{- 270 - 120}{130}

cos^{2} (θ) = \frac{2 - 3}{4}

What is the general solution for cos(θ)=(sqrt(2))/2 ,sin(θ)=-(sqrt(2))/2 ?
The general solution for cos(θ)=(sqrt(2))/2 ,sin(θ)=-(sqrt(2))/2 is No Solution for θ\in\mathbb{R}