What is the general solution for cos(2x)= 1/9 ?

The general solution for cos(2x)= 1/9 is x=(1.45945…)/2+pin,x=pi-(1.45945…)/2+pin

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cos(2x)= 1/9

cos (2 x) = \frac{1}{9}

x = \frac{1.45945 \dots}{2} + πn, x = π - \frac{1.45945 \dots}{2} + πn

Degrees

x = 41.81031 \dots^{\circ} + 18 0^{\circ} n, x = 138.18968 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 x) = \frac{1}{9}

Apply trig inverse properties

cos (2 x) = \frac{1}{9}

General solutions for

cos (2 x) = \frac{1}{9}

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

2 x = arccos (\frac{1}{9}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{9}) + 2 πn

2 x = arccos (\frac{1}{9}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{9}) + 2 πn

Solve

2 x = arccos (\frac{1}{9}) + 2 πn : x = \frac{arccos ( \frac{1}{9} )}{2} + πn

2 x = arccos (\frac{1}{9}) + 2 πn

Divide both sides by

2

2 x = arccos (\frac{1}{9}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( \frac{1}{9} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arccos ( \frac{1}{9} )}{2} + πn

x = \frac{arccos ( \frac{1}{9} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{1}{9}) + 2 πn : x = π - \frac{arccos ( \frac{1}{9} )}{2} + πn

2 x = 2 π - arccos (\frac{1}{9}) + 2 πn

Divide both sides by

2

2 x = 2 π - arccos (\frac{1}{9}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{1}{9} )}{2} + \frac{2 πn}{2}

Simplify

x = π - \frac{arccos ( \frac{1}{9} )}{2} + πn

x = π - \frac{arccos ( \frac{1}{9} )}{2} + πn

x = \frac{arccos ( \frac{1}{9} )}{2} + πn, x = π - \frac{arccos ( \frac{1}{9} )}{2} + πn

Show solutions in decimal form

x = \frac{1.45945 \dots}{2} + πn, x = π - \frac{1.45945 \dots}{2} + πn

\frac{( 1 )}{( cot ( θ ) )} = \frac{- ( 7 )}{( 25 )}

tan (θ) = \frac{- 1}{- 2}

sin (θ) = - \frac{1}{3}, \frac{3 π}{2} < θ < 2 π

4 cos (x) = 5 - 1

2 sin^{2} (θ) - 2 sin (θ) = 0

What is the general solution for cos(2x)= 1/9 ?
The general solution for cos(2x)= 1/9 is x=(1.45945…)/2+pin,x=pi-(1.45945…)/2+pin