What is the general solution for cos(2x)+7cos(x)+4=0 ?

The general solution for cos(2x)+7cos(x)+4=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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cos(2x)+7cos(x)+4=0

cos (2 x) + 7 cos (x) + 4 = 0

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

Solution steps

cos (2 x) + 7 cos (x) + 4 = 0

Rewrite using trig identities

3 + 2 cos^{2} (x) + 7 cos (x) = 0

Solve by substitution

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

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

cos (x) = - 3 :

No Solution

Combine all the solutions

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

2 sec (θ) - 2 = 0

8 cos (2 θ) = 24 cos (θ) - 16

x, y = sin (x^{2})

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

6 sec^{2} (x) - 6 = 0

What is the general solution for cos(2x)+7cos(x)+4=0 ?
The general solution for cos(2x)+7cos(x)+4=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin