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

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

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

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

x = \frac{π}{6} + \frac{2 πn}{3}

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Angle Difference identity:

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

= sin (4 x - x)

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}

sec (θ) = \frac{10}{6}

x, 2 cos (3 x) = - 3

x, 7 - 3 cos (x) = 9 sin^{2} (x)

4 sin (x) = 3 tan (x)

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

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