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

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

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

cos^{2} (2 x) = cos (4 x)

x = πn, x = \frac{π}{2} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

cos^{2} (2 x) = cos (4 x)

Subtract

cos (4 x)

from both sides

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

Rewrite using trig identities

cos^{2} (2 x) - cos (4 x)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

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

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

Simplify

cos (2 \cdot 2 x) + sin^{2} (2 x) - cos (4 x) : sin^{2} (2 x)

cos (2 \cdot 2 x) + sin^{2} (2 x) - cos (4 x)

Multiply the numbers:

2 \cdot 2 = 4

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

Add similar elements:

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

= sin^{2} (2 x)

= sin^{2} (2 x)

sin^{2} (2 x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (2 x) = 0

General solutions for

sin (2 x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

θ, tan (2 θ) = - 1

3 cos (z + 1.2) = 2

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

cos (\frac{x}{2}) = - 12

4 sin (β) + 2 = 3 + 2 sin (β)

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