What is the general solution for cos(2x)-5cos(x)-1=0 ?

The general solution for cos(2x)-5cos(x)-1=0 is x=1.92920…+2pin,x=-1.92920…+2pin

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cos(2x)-5cos(x)-1=0

cos (2 x) - 5 cos (x) - 1 = 0

x = 1.92920 \dots + 2 πn, x = - 1.92920 \dots + 2 πn

Solution steps

cos (2 x) - 5 cos (x) - 1 = 0

Rewrite using trig identities

- 2 + 2 cos^{2} (x) - 5 cos (x) = 0

Solve by substitution

cos (x) = \frac{5 + 41}{4}, cos (x) = \frac{5 - 41}{4}

cos (x) = \frac{5 + 41}{4} :

No Solution

cos (x) = \frac{5 - 41}{4} : x = arccos (\frac{5 - 41}{4}) + 2 πn, x = - arccos (\frac{5 - 41}{4}) + 2 πn

Combine all the solutions

x = arccos (\frac{5 - 41}{4}) + 2 πn, x = - arccos (\frac{5 - 41}{4}) + 2 πn

Show solutions in decimal form

x = 1.92920 \dots + 2 πn, x = - 1.92920 \dots + 2 πn

3 + 3 cos (θ) = 3

csc^{2} (x) = 3 csc (x) + 4

4 cot (2 θ) = 0

sin (6 0^{\circ}) = cos (x + 1 0^{\circ})

sin (2 x) = 682

What is the general solution for cos(2x)-5cos(x)-1=0 ?
The general solution for cos(2x)-5cos(x)-1=0 is x=1.92920…+2pin,x=-1.92920…+2pin