What is the general solution for picos(θ)=-pi/2 ?

The general solution for picos(θ)=-pi/2 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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picos(θ)=-pi/2

π cos (θ) = - \frac{π}{2}

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

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

π cos (θ) = - \frac{π}{2}

Divide both sides by

π

π cos (θ) = - \frac{π}{2}

Divide both sides by

π

\frac{π cos ( θ )}{π} = \frac{- \frac{π}{2}}{π}

Simplify

\frac{π cos ( θ )}{π} = \frac{- \frac{π}{2}}{π}

Simplify

\frac{π cos ( θ )}{π} : cos (θ)

\frac{π cos ( θ )}{π}

Cancel the common factor:

π

= cos (θ)

Simplify

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

\frac{- \frac{π}{2}}{π}

Apply the fraction rule:

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

= - \frac{\frac{π}{2}}{π}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

\frac{\frac{π}{2}}{π} = \frac{π}{2 π}

= - \frac{π}{2 π}

Cancel the common factor:

π

= - \frac{1}{2}

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{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{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

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

x, y = 1 + sin (\frac{π x}{4})

sin (x) + sin (\frac{x}{2}) = 0

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

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

cos^{2} (θ) - 3 sin (2 θ) + sin^{2} (θ) + 2 sin^{2} (2 θ) = 0

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