What is the general solution for 3+4cos(2x)+2cos(4x)=0 ?

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

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3+4cos(2x)+2cos(4x)=0

3 + 4 cos (2 x) + 2 cos (4 x) = 0

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

Solution steps

3 + 4 cos (2 x) + 2 cos (4 x) = 0

Rewrite using trig identities

1 + 4 cos (2 x) + 4 cos^{2} (2 x) = 0

Solve by substitution

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

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

Combine all the solutions

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

cos (θ) = - \frac{4}{5}, sin (2 θ), 18 0^{\circ} < θ < 27 0^{\circ}

2 sin (θ) + 1 = 2

2 - 2 sin^{2} (3 x) = 1

4 cos^{2} (x) = 5 - 4 sin (x), 0 \leq x \leq 2 π

0 = 6 sin (x)

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