What is the general solution for -6cos(2x)+23cos(x)-11=0 ?

The general solution for -6cos(2x)+23cos(x)-11=0 is x=1.31811…+2pin,x=2pi-1.31811…+2pin

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-6cos(2x)+23cos(x)-11=0

- 6 cos (2 x) + 23 cos (x) - 11 = 0

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

Solution steps

- 6 cos (2 x) + 23 cos (x) - 11 = 0

Rewrite using trig identities

- 5 - 12 cos^{2} (x) + 23 cos (x) = 0

Solve by substitution

cos (x) = \frac{1}{4}, cos (x) = \frac{5}{3}

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

cos (x) = \frac{5}{3} :

No Solution

Combine all the solutions

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

Show solutions in decimal form

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn

sin^{2} (x + 4) = 1

\frac{178.9}{sin ( 12 5 ^{\circ} )} = \frac{125}{sin ( x )}

cos (θ) = \frac{3}{8}

arccos (x) - arcsin (x) = arccos (\frac{3}{2})

- 5 tan (θ) - 12 = tan (θ) - 8

What is the general solution for -6cos(2x)+23cos(x)-11=0 ?
The general solution for -6cos(2x)+23cos(x)-11=0 is x=1.31811…+2pin,x=2pi-1.31811…+2pin