What is the general solution for 0=a(1+cos(2θ)) ?

The general solution for 0=a(1+cos(2θ)) is θ= pi/2+pin

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0=a(1+cos(2θ))

0 = a (1 + cos (2 θ))

θ = \frac{π}{2} + πn

Solution steps

0 = a (1 + cos (2 θ))

Switch sides

a (1 + cos (2 θ)) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

1 + cos (2 θ) = 0

Move

1

to the right side

cos (2 θ) = - 1

General solutions for

cos (2 θ) = - 1

2 θ = π + 2 πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

4 sin (2 x) = 3 cos (x)

sin (b) = (\frac{250 km}{590 km}) sin (2 3^{\circ}), 8 2^{\circ}

sin (x) = \frac{30}{60}

4 cos (z) + 5 = 0

y, sin (x^{2} y^{2}) = x

What is the general solution for 0=a(1+cos(2θ)) ?
The general solution for 0=a(1+cos(2θ)) is θ= pi/2+pin