What is the general solution for (1+cos(4x))sin(4x)=2cos^2(2x) ?

The general solution for (1+cos(4x))sin(4x)=2cos^2(2x) is x= pi/4+pin,x=(3pi)/4+pin,x= pi/8+(pin)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1+cos(4x))sin(4x)=2cos^2(2x)

Solution

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

Solution

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn, x = \frac{π}{8} + \frac{πn}{2}

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n, x = 22. 5^{\circ} + 9 0^{\circ} n

Solution steps

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

Subtract

2 cos^{2} (2 x)

from both sides

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

Rewrite using trig identities

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

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

Simplify

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

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

Group like terms

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

1 - 1 = 0

= 2 cos^{2} (2 x)

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

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

Factor

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

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

Factor out common term

2 cos^{2} (2 x)

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

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

Solving each part separately

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

cos^{2} (2 x) = 0 : x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

cos^{2} (2 x) = 0

Apply rule

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

cos (2 x) = 0

General solutions for

cos (2 x) = 0

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

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

Solve

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{π}{2}}{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}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solve

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{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{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

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

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

sin (4 x) - 1 = 0 : x = \frac{π}{8} + \frac{πn}{2}

sin (4 x) - 1 = 0

Move

1

to the right side

sin (4 x) - 1 = 0

Add

1

to both sides

sin (4 x) - 1 + 1 = 0 + 1

Simplify

sin (4 x) = 1

sin (4 x) = 1

General solutions for

sin (4 x) = 1

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}

4 x = \frac{π}{2} + 2 πn

4 x = \frac{π}{2} + 2 πn

Solve

4 x = \frac{π}{2} + 2 πn : x = \frac{π}{8} + \frac{πn}{2}

4 x = \frac{π}{2} + 2 πn

Divide both sides by

4

4 x = \frac{π}{2} + 2 πn

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{π}{2}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{π}{2}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4}

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn, x = \frac{π}{8} + \frac{πn}{2}

Popular Examples

(5sin(x)+6)/(sin(x))=17

\frac{5 sin ( x ) + 6}{sin ( x )} = 17

tan^2(x)sec(x)-1=0

tan^{2} (x) sec (x) - 1 = 0

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

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

(1+sin^2(x))=(4cos(x))^2

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

cos(x)=53

cos (x) = 53

Frequently Asked Questions (FAQ)

What is the general solution for (1+cos(4x))sin(4x)=2cos^2(2x) ?
The general solution for (1+cos(4x))sin(4x)=2cos^2(2x) is x= pi/4+pin,x=(3pi)/4+pin,x= pi/8+(pin)/2