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

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

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-pi/2 =picos(θ)

- \frac{π}{2} = π cos (θ)

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

- \frac{π}{2} = π cos (θ)

Switch sides

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

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

sin (x) tan (x) + 1 = sin (x) + tan (x)

\frac{2 \cdot 40}{80 + 20} = tan (2 x)

x, sec (x + 1 0^{\circ}) = 5.759

tan (x) cot (x) + 2 cot (x) = 0

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