What is the general solution for 3cos(2x)-1=0 ?

The general solution for 3cos(2x)-1=0 is x=(1.23095…)/2+pin,x=pi-(1.23095…)/2+pin

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3cos(2x)-1=0

3 cos (2 x) - 1 = 0

x = \frac{1.23095 \dots}{2} + πn, x = π - \frac{1.23095 \dots}{2} + πn

Degrees

x = 35.26438 \dots^{\circ} + 18 0^{\circ} n, x = 144.73561 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 cos (2 x) - 1 = 0

Move

1

to the right side

3 cos (2 x) - 1 = 0

Add

1

to both sides

3 cos (2 x) - 1 + 1 = 0 + 1

Simplify

3 cos (2 x) = 1

3 cos (2 x) = 1

Divide both sides by

3

3 cos (2 x) = 1

Divide both sides by

3

\frac{3 cos ( 2 x )}{3} = \frac{1}{3}

Simplify

cos (2 x) = \frac{1}{3}

cos (2 x) = \frac{1}{3}

Apply trig inverse properties

cos (2 x) = \frac{1}{3}

General solutions for

cos (2 x) = \frac{1}{3}

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

2 x = arccos (\frac{1}{3}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{3}) + 2 πn

2 x = arccos (\frac{1}{3}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{3}) + 2 πn

Solve

2 x = arccos (\frac{1}{3}) + 2 πn : x = \frac{arccos ( \frac{1}{3} )}{2} + πn

2 x = arccos (\frac{1}{3}) + 2 πn

Divide both sides by

2

2 x = arccos (\frac{1}{3}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arccos ( \frac{1}{3} )}{2} + πn

x = \frac{arccos ( \frac{1}{3} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{1}{3}) + 2 πn : x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

2 x = 2 π - arccos (\frac{1}{3}) + 2 πn

Divide both sides by

2

2 x = 2 π - arccos (\frac{1}{3}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{3} )}{2} + \frac{2 πn}{2}

Simplify

x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

x = \frac{arccos ( \frac{1}{3} )}{2} + πn, x = π - \frac{arccos ( \frac{1}{3} )}{2} + πn

Show solutions in decimal form

x = \frac{1.23095 \dots}{2} + πn, x = π - \frac{1.23095 \dots}{2} + πn

2 cos^{2} (x) = 2

0 = - 2 sin (t) - 4 cos (2 t)

cos (θ) + sin (θ + \frac{π}{2}) = 1, 0 \leq θ < 2 π

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

42 cos (x) - 8 sin (x) cos (x) = 0

What is the general solution for 3cos(2x)-1=0 ?
The general solution for 3cos(2x)-1=0 is x=(1.23095…)/2+pin,x=pi-(1.23095…)/2+pin