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

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

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

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

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

Solution steps

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

Solve by substitution

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

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

cos (x) = - 1 : x = π + 2 πn

Combine all the solutions

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

sin (A) = \frac{6.347}{12.7}

sec (x) = 3 cos (x) - tan (x)

arctan (x^{2} - 2 x) = 0

sin (x^{\circ}) = \frac{6.57 \cdot 1 0 ^{- 7}}{1.8 \cdot 1 0 ^{- 6}}

3 sin^{2} (x) + cos^{2} (x) = 3 sin (x)

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