What is the general solution for 0=3cos(θ) ?

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

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0=3cos(θ)

0 = 3 cos (θ)

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

0 = 3 cos (θ)

Switch sides

3 cos (θ) = 0

Divide both sides by

3

3 cos (θ) = 0

Divide both sides by

3

\frac{3 cos ( θ )}{3} = \frac{0}{3}

Simplify

cos (θ) = 0

cos (θ) = 0

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

1 - 3 sin^{2} (x) = cos^{2} (x) - sin^{2} (x)

sin^{2} (θ) - sin (θ) + 1 = cos^{2} (θ)

1 = l cos (x)

tan (2 x) = cos (2 x), 0 \leq x \leq π

5 sin (2 x) = 5 cos (x), 0 \leq x \leq 2 π

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