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

Degrees

x = 37.76124 \dots^{\circ} + 18 0^{\circ} n, x = 142.23875 \dots^{\circ} + 18 0^{\circ} n

Solution steps

1 - 4 cos (2 x) = 0

Move

1

to the right side

1 - 4 cos (2 x) = 0

Subtract

1

from both sides

1 - 4 cos (2 x) - 1 = 0 - 1

Simplify

- 4 cos (2 x) = - 1

- 4 cos (2 x) = - 1

Divide both sides by

- 4

- 4 cos (2 x) = - 1

Divide both sides by

- 4

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

2 x = arccos (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 x = arccos (\frac{1}{4}) + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

x = π - \frac{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