What is the general solution for -10cos(2x)-32cos(x)-22=0 ?

The general solution for -10cos(2x)-32cos(x)-22=0 is x=pi+2pin,x=2.21429…+2pin,x=-2.21429…+2pin

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-10cos(2x)-32cos(x)-22=0

- 10 cos (2 x) - 32 cos (x) - 22 = 0

x = π + 2 πn, x = 2.21429 \dots + 2 πn, x = - 2.21429 \dots + 2 πn

Solution steps

- 10 cos (2 x) - 32 cos (x) - 22 = 0

Rewrite using trig identities

- 12 - 20 cos^{2} (x) - 32 cos (x) = 0

Solve by substitution

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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - \frac{3}{5} : x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

Combine all the solutions

x = π + 2 πn, x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

Show solutions in decimal form

x = π + 2 πn, x = 2.21429 \dots + 2 πn, x = - 2.21429 \dots + 2 πn

θ, sin (θ) = - \frac{3}{4}

tan (x) \cdot sec (x) = 2

cos (2 x + \frac{π}{3}) = sin (2 x)

cot (2 t) = 0

tan (θ) = cot (θ), 0 \leq θ < 2 π

What is the general solution for -10cos(2x)-32cos(x)-22=0 ?
The general solution for -10cos(2x)-32cos(x)-22=0 is x=pi+2pin,x=2.21429…+2pin,x=-2.21429…+2pin