What is the general solution for 2cos(2x)+8cos(x)+5=0 ?

The general solution for 2cos(2x)+8cos(x)+5=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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2cos(2x)+8cos(x)+5=0

2 cos (2 x) + 8 cos (x) + 5 = 0

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solution steps

2 cos (2 x) + 8 cos (x) + 5 = 0

Rewrite using trig identities

3 + 4 cos^{2} (x) + 8 cos (x) = 0

Solve by substitution

cos (x) = - \frac{1}{2}, cos (x) = - \frac{3}{2}

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

cos (x) = - \frac{3}{2} :

No Solution

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sin (x) \cdot tan (x) = 3 \cdot sin (x)

csc (x) = 2.6

2 sin^{2} (t) + 3 sin (t) + 1 = 0

cos (x) = - 0.6060400000000 \dots

sin (x) cot (x) = 0.5

What is the general solution for 2cos(2x)+8cos(x)+5=0 ?
The general solution for 2cos(2x)+8cos(x)+5=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin