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

The general solution for cos(θ)=-(1/(sqrt(2))) is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin

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cos(θ)=-(1/(sqrt(2)))

cos (θ) = - (\frac{1}{2})

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

Solution steps

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{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos (θ) = 2 cos (θ) - 1

2 cos^{2} (2 β) + cos (2 β) - 1 = 0, - 36 0^{\circ} \leq β \leq 36 0^{\circ}

cos (2 x) = \frac{3}{7}

0 = 1 - 3 cos (θ)

(tan (\frac{3 a}{2})) tan (\frac{a}{2}) = 3

What is the general solution for cos(θ)=-(1/(sqrt(2))) ?
The general solution for cos(θ)=-(1/(sqrt(2))) is θ=(3pi)/4+2pin,θ=(5pi)/4+2pin