What is the general solution for 7cos(9x)-1=1 ?

The general solution for 7cos(9x)-1=1 is x=(1.28104…)/9+(2pin)/9 ,x=(2pi)/9-(1.28104…)/9+(2pin)/9

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7cos(9x)-1=1

7 cos (9 x) - 1 = 1

x = \frac{1.28104 \dots}{9} + \frac{2 πn}{9}, x = \frac{2 π}{9} - \frac{1.28104 \dots}{9} + \frac{2 πn}{9}

Solution steps

7 cos (9 x) - 1 = 1

Move

1

to the right side

7 cos (9 x) = 2

Divide both sides by

7

cos (9 x) = \frac{2}{7}

Apply trig inverse properties

9 x = arccos (\frac{2}{7}) + 2 πn, 9 x = 2 π - arccos (\frac{2}{7}) + 2 πn

Solve

9 x = arccos (\frac{2}{7}) + 2 πn : x = \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

Solve

9 x = 2 π - arccos (\frac{2}{7}) + 2 πn : x = \frac{2 π}{9} - \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

x = \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}, x = \frac{2 π}{9} - \frac{arccos ( \frac{2}{7} )}{9} + \frac{2 πn}{9}

Show solutions in decimal form

x = \frac{1.28104 \dots}{9} + \frac{2 πn}{9}, x = \frac{2 π}{9} - \frac{1.28104 \dots}{9} + \frac{2 πn}{9}

cos (22. 5^{\circ})

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

sin (- \frac{2 π}{3})

arcsin (- 1)

tan (1)

What is the general solution for 7cos(9x)-1=1 ?
The general solution for 7cos(9x)-1=1 is x=(1.28104…)/9+(2pin)/9 ,x=(2pi)/9-(1.28104…)/9+(2pin)/9