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

The general solution for cos(2x)+cos(x)=1 is x=0.67488…+2pin,x=2pi-0.67488…+2pin

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cos(2x)+cos(x)=1

cos (2 x) + cos (x) = 1

x = 0.67488 \dots + 2 πn, x = 2 π - 0.67488 \dots + 2 πn

Solution steps

cos (2 x) + cos (x) = 1

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = \frac{- 1 + 17}{4}, cos (x) = \frac{- 1 - 17}{4}

cos (x) = \frac{- 1 + 17}{4} : x = arccos (\frac{- 1 + 17}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 17}{4}) + 2 πn

cos (x) = \frac{- 1 - 17}{4} :

No Solution

Combine all the solutions

x = arccos (\frac{- 1 + 17}{4}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 17}{4}) + 2 πn

Show solutions in decimal form

x = 0.67488 \dots + 2 πn, x = 2 π - 0.67488 \dots + 2 πn

tan (x) sec (x) = - 2 tan (x)

1 + sin^{2} (x) = 2

sin (\frac{3 π}{2} + x) + sin (\frac{3 π}{2} + x) = - 2

cos (x) tan (x) = 0

cos^{2} (x) + 4 cos (x) + 3 = 0

What is the general solution for cos(2x)+cos(x)=1 ?
The general solution for cos(2x)+cos(x)=1 is x=0.67488…+2pin,x=2pi-0.67488…+2pin