What is the general solution for 40+40cos((2pix)/9)=50 ?

The general solution for 40+40cos((2pix)/9)=50 is x=(9*1.31811…)/(2pi)+9n,x=9-(9*1.31811…)/(2pi)+9n

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40+40cos((2pix)/9)=50

40 + 40 cos (\frac{2 π x}{9}) = 50

x = \frac{9 \cdot 1.31811 \dots}{2 π} + 9 n, x = 9 - \frac{9 \cdot 1.31811 \dots}{2 π} + 9 n

Solution steps

40 + 40 cos (\frac{2 π x}{9}) = 50

Move

40

to the right side

40 cos (\frac{2 π x}{9}) = 10

Divide both sides by

40

cos (\frac{2 π x}{9}) = \frac{1}{4}

Apply trig inverse properties

\frac{2 π x}{9} = arccos (\frac{1}{4}) + 2 πn, \frac{2 π x}{9} = 2 π - arccos (\frac{1}{4}) + 2 πn

Solve

\frac{2 π x}{9} = arccos (\frac{1}{4}) + 2 πn : x = \frac{9 arccos ( \frac{1}{4} )}{2 π} + 9 n

Solve

\frac{2 π x}{9} = 2 π - arccos (\frac{1}{4}) + 2 πn : x = 9 - \frac{9 arccos ( \frac{1}{4} )}{2 π} + 9 n

x = \frac{9 arccos ( \frac{1}{4} )}{2 π} + 9 n, x = 9 - \frac{9 arccos ( \frac{1}{4} )}{2 π} + 9 n

Show solutions in decimal form

x = \frac{9 \cdot 1.31811 \dots}{2 π} + 9 n, x = 9 - \frac{9 \cdot 1.31811 \dots}{2 π} + 9 n

tan (x) + sec (x) - 1 = 0

sin (x) (sin (x) - 1) = 3 cos^{2} (x)

1 = 1 + sec (x)

x, cos^{2} (x) = 4 p + 6

sin (x) = - 1, 0 < x < 2 π

What is the general solution for 40+40cos((2pix)/9)=50 ?
The general solution for 40+40cos((2pix)/9)=50 is x=(9*1.31811…)/(2pi)+9n,x=9-(9*1.31811…)/(2pi)+9n