What is the general solution for 3cos(2x)+4cos(4x)-5=0 ?

The general solution for 3cos(2x)+4cos(4x)-5=0 is x=(0.47431…)/2+pin,x=pi-(0.47431…)/2+pin

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3cos(2x)+4cos(4x)-5=0

3 cos (2 x) + 4 cos (4 x) - 5 = 0

x = \frac{0.47431 \dots}{2} + πn, x = π - \frac{0.47431 \dots}{2} + πn

Solution steps

3 cos (2 x) + 4 cos (4 x) - 5 = 0

Rewrite using trig identities

- 9 + 3 cos (2 x) + 8 cos^{2} (2 x) = 0

Solve by substitution

cos (2 x) = \frac{- 3 + 3 33}{16}, cos (2 x) = \frac{- 3 - 3 33}{16}

cos (2 x) = \frac{- 3 + 3 33}{16} : x = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn, x = π - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

cos (2 x) = \frac{- 3 - 3 33}{16} :

No Solution

Combine all the solutions

x = \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn, x = π - \frac{arccos ( \frac{- 3 + 3 33}{16} )}{2} + πn

Show solutions in decimal form

x = \frac{0.47431 \dots}{2} + πn, x = π - \frac{0.47431 \dots}{2} + πn

cos (2 θ) = \frac{20}{29}

- 6 sin (3 x) = 0

sin (x) = \frac{2}{4}

2 tan (2 x) - 6 = 0

8 sin (θ) + 15 cos (θ) = 18

What is the general solution for 3cos(2x)+4cos(4x)-5=0 ?
The general solution for 3cos(2x)+4cos(4x)-5=0 is x=(0.47431…)/2+pin,x=pi-(0.47431…)/2+pin