What is the general solution for 0=cos^2(x)+cos(x)-1 ?

The general solution for 0=cos^2(x)+cos(x)-1 is x=0.90455…+2pin,x=2pi-0.90455…+2pin

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

0 = cos^{2} (x) + cos (x) - 1

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

Solution steps

0 = cos^{2} (x) + cos (x) - 1

Switch sides

cos^{2} (x) + cos (x) - 1 = 0

Solve by substitution

cos (x) = \frac{- 1 + 5}{2}, cos (x) = \frac{- 1 - 5}{2}

cos (x) = \frac{- 1 + 5}{2} : x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{- 1 - 5}{2} :

No Solution

Combine all the solutions

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

\frac{tan ( x ) ( - tan ^{2} ( x ) + 3 )}{1 - tan ^{2} ( x )} = 0

2 sin (x) - cos (x) = 0.6

2 sin (x) - cos (x) = 0.5

cos (x) - tan (x) = 0

2 sin (x) - cos (x) = 0.2

What is the general solution for 0=cos^2(x)+cos(x)-1 ?
The general solution for 0=cos^2(x)+cos(x)-1 is x=0.90455…+2pin,x=2pi-0.90455…+2pin