What is the general solution for 2cos(2x)= 1/2 ?

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

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2cos(2x)= 1/2

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

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

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 cos ( 2 x )}{2} : cos (2 x)

\frac{2 cos ( 2 x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= cos (2 x)

Simplify

\frac{\frac{1}{2}}{2} : \frac{1}{4}

\frac{\frac{1}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{1}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1}{4}

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

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{sin ( 7 3 ^{\circ} )}{34} = \frac{sin ( d )}{29}

θ, \frac{9 0 ^{2}}{250 ( 32.2 )} = tan (θ)

sin (θ) = 2 sin (θ) cos (θ)

x, arccos (e^{- x}) = \frac{π}{3}

2 cos^{2} (4 x + 1) = 0

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