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

The general solution for 2cos(4x)+3=4cos(2x) is x= pi/6+pin,x=(5pi)/6+pin

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

Solution

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

Solution

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

4 cos (2 x)

from both sides

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

Rewrite using trig identities

3 + 2 cos (4 x) - 4 cos (2 x)

cos (4 x) = 2 cos^{2} (2 x) - 1

cos (4 x)

Rewrite as

= cos (2 \cdot 2 x)

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

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

= 2 cos^{2} (2 x) - 1

= 3 + 2 (2 cos^{2} (2 x) - 1) - 4 cos (2 x)

Simplify

3 + 2 (2 cos^{2} (2 x) - 1) - 4 cos (2 x) : 4 cos^{2} (2 x) - 4 cos (2 x) + 1

3 + 2 (2 cos^{2} (2 x) - 1) - 4 cos (2 x)

Expand

2 (2 cos^{2} (2 x) - 1) : 4 cos^{2} (2 x) - 2

2 (2 cos^{2} (2 x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 2 cos^{2} (2 x), c = 1

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

Simplify

2 \cdot 2 cos^{2} (2 x) - 2 \cdot 1 : 4 cos^{2} (2 x) - 2

2 \cdot 2 cos^{2} (2 x) - 2 \cdot 1

Multiply the numbers:

2 \cdot 2 = 4

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

= 3 + 4 cos^{2} (2 x) - 2 - 4 cos (2 x)

Simplify

3 + 4 cos^{2} (2 x) - 2 - 4 cos (2 x) : 4 cos^{2} (2 x) - 4 cos (2 x) + 1

3 + 4 cos^{2} (2 x) - 2 - 4 cos (2 x)

Group like terms

= 4 cos^{2} (2 x) - 4 cos (2 x) + 3 - 2

Add/Subtract the numbers:

3 - 2 = 1

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

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

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

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

Solve by substitution

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

Let:

cos (2 x) = u

1 - 4 u + 4 u^{2} = 0

1 - 4 u + 4 u^{2} = 0 : u = \frac{1}{2}

1 - 4 u + 4 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

4 u^{2} - 4 u + 1 = 0

Solve with the quadratic formula

4 u^{2} - 4 u + 1 = 0

Quadratic Equation Formula:

For

a = 4, b = - 4, c = 1

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

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

(- 4)^{2} - 4 \cdot 4 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 4}

u = \frac{- ( - 4 )}{2 \cdot 4}

\frac{- ( - 4 )}{2 \cdot 4} = \frac{1}{2}

\frac{- ( - 4 )}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{4}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = cos (2 x)

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

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

cos (2 x) = \frac{1}{2} : x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

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

General solutions for

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

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solve

2 x = \frac{π}{3} + 2 πn : x = \frac{π}{6} + πn

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

Solve

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{6} + πn

2 x = \frac{5 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Combine all the solutions

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(4x)+3=4cos(2x) ?
The general solution for 2cos(4x)+3=4cos(2x) is x= pi/6+pin,x=(5pi)/6+pin