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

The general solution for 2cos(2x)=4cos(x) is x=1.94553…+2pin,x=-1.94553…+2pin

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2cos(2x)=4cos(x)

2 cos (2 x) = 4 cos (x)

x = 1.94553 \dots + 2 πn, x = - 1.94553 \dots + 2 πn

Solution steps

2 cos (2 x) = 4 cos (x)

Subtract

4 cos (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = \frac{1 + 3}{2} :

No Solution

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

Combine all the solutions

x = arccos (\frac{1 - 3}{2}) + 2 πn, x = - arccos (\frac{1 - 3}{2}) + 2 πn

Show solutions in decimal form

x = 1.94553 \dots + 2 πn, x = - 1.94553 \dots + 2 πn

\frac{sin ( 4 2 ^{\circ} )}{22} = \frac{sin ( B )}{12}

sin (θ) = \frac{16}{14}

tan (θ) = \frac{83}{47}

x, 2 sin (x) - cos (x) sin (x) = 0

5 = 5 sin (4 θ)

What is the general solution for 2cos(2x)=4cos(x) ?
The general solution for 2cos(2x)=4cos(x) is x=1.94553…+2pin,x=-1.94553…+2pin