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

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

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solvefor θ,cos(θ)=(-1sqrt(2))/2

solve for

θ, cos (θ) = \frac{- 1 2}{2}

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

Solution steps

cos (θ) = \frac{- 1 \cdot 2}{2}

Simplify

\frac{- 1 \cdot 2}{2} : - \frac{2}{2}

\frac{- 1 \cdot 2}{2}

Multiply:

1 \cdot 2 = 2

= \frac{- 2}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \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^{2} (x) - sin (x) + 1 = 0

sin (2 x) = cos (x - 1 5^{\circ})

(1 + cos (4 x)) sin (4 x) = cos^{2} (2 x)

arcsin (\frac{3}{5}) + arccos (x) = π

cos (θ) = \frac{6}{7}

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