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

The general solution for [sin(4x)cos(x)-cos(4x)sin(x)]^2=1 is x= pi/2+(2pin)/3 ,x= pi/6+(2pin)/3

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Popular >

[sin(4x)cos(x)-cos(4x)sin(x)]^2=1

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 12 0^{\circ} n, x = 3 0^{\circ} + 12 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Factor

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

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

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Refine

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

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

Solving each part separately

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

sin (4 x) cos (x) - cos (4 x) sin (x) + 1 = 0 : x = \frac{π}{2} + \frac{2 πn}{3}

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

Rewrite using trig identities

sin (4 x) cos (x) - cos (4 x) sin (x) + 1

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= 1 + sin (4 x - x)

1 + sin (4 x - x) = 0

Move

1

to the right side

1 + sin (4 x - x) = 0

Subtract

1

from both sides

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

Simplify

sin (4 x - x) = - 1

sin (4 x - x) = - 1

General solutions for

sin (4 x - 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 - x = \frac{3 π}{2} + 2 πn

4 x - x = \frac{3 π}{2} + 2 πn

Solve

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

4 x - x = \frac{3 π}{2} + 2 πn

Add similar elements:

4 x - x = 3 x

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

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

\frac{\frac{3 π}{2}}{3} = \frac{π}{2}

\frac{\frac{3 π}{2}}{3}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6}

Cancel the common factor:

3

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

sin (4 x) cos (x) - cos (4 x) sin (x) - 1 = 0 : x = \frac{π}{6} + \frac{2 πn}{3}

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

Rewrite using trig identities

sin (4 x) cos (x) - cos (4 x) sin (x) - 1

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= - 1 + sin (4 x - x)

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

Move

1

to the right side

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

Add

1

to both sides

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

Simplify

sin (4 x - x) = 1

sin (4 x - x) = 1

General solutions for

sin (4 x - 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 - x = \frac{π}{2} + 2 πn

4 x - x = \frac{π}{2} + 2 πn

Solve

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

4 x - x = \frac{π}{2} + 2 πn

Add similar elements:

4 x - x = 3 x

3 x = \frac{π}{2} + 2 πn

Divide both sides by

3

3 x = \frac{π}{2} + 2 πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

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

\frac{\frac{π}{2}}{3} = \frac{π}{6}

\frac{\frac{π}{2}}{3}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

x = \frac{π}{6} + \frac{2 πn}{3}

Combine all the solutions

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

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cos (θ) = \frac{202}{331 \cdot 2 33}

5sin(θ)+12cos(θ)=13

5 sin (θ) + 12 cos (θ) = 13

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

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

cot(x)= 5/12 ,pi<x<(3pi)/2

cot (x) = \frac{5}{12}, π < x < \frac{3 π}{2}

tan(x)= 1/5 ,tan(x+pi/2)

tan (x) = \frac{1}{5}, tan (x + \frac{π}{2})

Frequently Asked Questions (FAQ)

What is the general solution for [sin(4x)cos(x)-cos(4x)sin(x)]^2=1 ?
The general solution for [sin(4x)cos(x)-cos(4x)sin(x)]^2=1 is x= pi/2+(2pin)/3 ,x= pi/6+(2pin)/3