What is the general solution for 1-4cos(2x)=0 ?

The general solution for 1-4cos(2x)=0 is x=(1.31811…)/2+pin,x=pi-(1.31811…)/2+pin

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1-4cos(2x)=0

1 - 4 cos (2 x) = 0

x = \frac{1.31811 \dots}{2} + πn, x = π - \frac{1.31811 \dots}{2} + πn

Solution steps

1 - 4 cos (2 x) = 0

Move

1

to the right side

- 4 cos (2 x) = - 1

Divide both sides by

- 4

cos (2 x) = \frac{1}{4}

Apply trig inverse properties

2 x = arccos (\frac{1}{4}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{4}) + 2 πn

Solve

2 x = arccos (\frac{1}{4}) + 2 πn : x = \frac{arccos ( \frac{1}{4} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{1}{4}) + 2 πn : x = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

x = \frac{arccos ( \frac{1}{4} )}{2} + πn, x = π - \frac{arccos ( \frac{1}{4} )}{2} + πn

Show solutions in decimal form

x = \frac{1.31811 \dots}{2} + πn, x = π - \frac{1.31811 \dots}{2} + πn

\frac{1}{2} = cos (θ)

tan^{2} (x) + 5 tan (x) = 0

1 - sinh^{2} (x) = 0

sin (y + 1 0^{\circ}) = cos (y + 2 0^{\circ})

tan (x) = - 1, 7

What is the general solution for 1-4cos(2x)=0 ?
The general solution for 1-4cos(2x)=0 is x=(1.31811…)/2+pin,x=pi-(1.31811…)/2+pin