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

The general solution for sin(θ)=(-1)/(sqrt(2)) is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin

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sin(θ)=(-1)/(sqrt(2))

sin (θ) = \frac{- 1}{2}

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Solution steps

sin (θ) = \frac{- 1}{2}

Simplify

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

\frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

Rationalize

- \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}

= - \frac{2}{2}

sin (θ) = - \frac{2}{2}

General solutions for

sin (θ) = - \frac{2}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

(sin (a) + cos (a))^{2} = 1 + 2, a cos (a)

4 sin (z) cos^{2} (z) = 3 sin (z)

3 (tan (x) + sin^{2} (x)) - 4 = 4 tan (x) - 3 cos^{2} (x)

\frac{1}{3} = tan (θ)

(sec^{(2)} (θ)) (cos (θ)) = 2 cos (θ)

What is the general solution for sin(θ)=(-1)/(sqrt(2)) ?
The general solution for sin(θ)=(-1)/(sqrt(2)) is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin