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

The general solution for cos(2x)-7cos(x)+4=0 is x= pi/3+2pin,x=(5pi)/3+2pin

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

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

x = \frac{π}{3} + 2 πn, x = \frac{5 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) = 3, cos (x) = \frac{1}{2}

cos (x) = 3 :

No Solution

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

\frac{87.17}{sin ( 12 0 ^{\circ} )} = \frac{40}{sin ( x )}

23 tan (x) - 2 = 0

2 cos (x) 2 = 0

cos^{2} (x) - 6 cos (x) = - 9

cos (x) - 3 = 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= pi/3+2pin,x=(5pi)/3+2pin