What is the general solution for (1+2cos(2x))^2=0 ?

The general solution for (1+2cos(2x))^2=0 is x= pi/3+pin,x=(2pi)/3+pin

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(1+2cos(2x))^2=0

(1 + 2 cos (2 x))^{2} = 0

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

Solution steps

(1 + 2 cos (2 x))^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

1 + 2 cos (2 x) = 0 : cos (2 x) = - \frac{1}{2}

The solution to the quadratic equation is:

cos (2 x) = - \frac{1}{2}

General solutions for

cos (2 x) = - \frac{1}{2}

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

Solve

2 x = \frac{2 π}{3} + 2 πn : x = \frac{π}{3} + πn

Solve

2 x = \frac{4 π}{3} + 2 πn : x = \frac{2 π}{3} + πn

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

2 + 3 tan (x) = 3

(sin (x) - 4) (sin (x) - 1) = 0

tan (2 x) = 0.6

2 sin (3 θ) = - 1

2 + 2 cos (θ) = 2 - 2 cos (θ)

What is the general solution for (1+2cos(2x))^2=0 ?
The general solution for (1+2cos(2x))^2=0 is x= pi/3+pin,x=(2pi)/3+pin