What is the general solution for cos(θ)=0,0<= θ<= 2pi ?

The general solution for cos(θ)=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2

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cos(θ)=0,0<= θ<= 2pi

cos (θ) = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Degrees

θ = 9 0^{\circ}, θ = 27 0^{\circ}

Solution steps

cos (θ) = 0, 0 \leq θ \leq 2 π

General solutions for

cos (θ) = 0

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

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

- 2 sin (x) - 2 + 4 sin^{2} (x) = 0

cos (x) = - \frac{3}{5}, π < x < \frac{3 π}{2}, sin (\frac{x}{2})

3 0^{\circ} = arccos (\frac{6}{x ( 3 )})

sin (θ) = 0.18421

\frac{6}{3 tan ( θ ) + 1} = 5

What is the general solution for cos(θ)=0,0<= θ<= 2pi ?
The general solution for cos(θ)=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2