What is the general solution for cos(2x)= 9/41 ?

The general solution for cos(2x)= 9/41 is x=(1.34948…)/2+pin,x=pi-(1.34948…)/2+pin

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cos(2x)= 9/41

cos (2 x) = \frac{9}{41}

x = \frac{1.34948 \dots}{2} + πn, x = π - \frac{1.34948 \dots}{2} + πn

Degrees

x = 38.65980 \dots^{\circ} + 18 0^{\circ} n, x = 141.34019 \dots^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 x) = \frac{9}{41}

Apply trig inverse properties

cos (2 x) = \frac{9}{41}

General solutions for

cos (2 x) = \frac{9}{41}

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

2 x = arccos (\frac{9}{41}) + 2 πn, 2 x = 2 π - arccos (\frac{9}{41}) + 2 πn

2 x = arccos (\frac{9}{41}) + 2 πn, 2 x = 2 π - arccos (\frac{9}{41}) + 2 πn

Solve

2 x = arccos (\frac{9}{41}) + 2 πn : x = \frac{arccos ( \frac{9}{41} )}{2} + πn

2 x = arccos (\frac{9}{41}) + 2 πn

Divide both sides by

2

2 x = arccos (\frac{9}{41}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( \frac{9}{41} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arccos ( \frac{9}{41} )}{2} + πn

x = \frac{arccos ( \frac{9}{41} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{9}{41}) + 2 πn : x = π - \frac{arccos ( \frac{9}{41} )}{2} + πn

2 x = 2 π - arccos (\frac{9}{41}) + 2 πn

Divide both sides by

2

2 x = 2 π - arccos (\frac{9}{41}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{9}{41} )}{2} + \frac{2 πn}{2}

Simplify

x = π - \frac{arccos ( \frac{9}{41} )}{2} + πn

x = π - \frac{arccos ( \frac{9}{41} )}{2} + πn

x = \frac{arccos ( \frac{9}{41} )}{2} + πn, x = π - \frac{arccos ( \frac{9}{41} )}{2} + πn

Show solutions in decimal form

x = \frac{1.34948 \dots}{2} + πn, x = π - \frac{1.34948 \dots}{2} + πn

sec (5 x) = \frac{2}{3}

cos (a) = - \frac{4}{9}, \frac{π}{2} \leq a \leq π

20 sin^{2} (x) - 26 sin (x) + 8 = 0

1.5 = 1 + 0.6 cos (\frac{π t}{2})

tan (x) csc (x) = cos (2)

What is the general solution for cos(2x)= 9/41 ?
The general solution for cos(2x)= 9/41 is x=(1.34948…)/2+pin,x=pi-(1.34948…)/2+pin