What is the general solution for sin(3x)cos(x)+cos(3x)sin(x)=1 ?

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

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sin(3x)cos(x)+cos(3x)sin(x)=1

sin (3 x) cos (x) + cos (3 x) sin (x) = 1

x = \frac{π}{8} + \frac{πn}{2}

Degrees

x = 22. 5^{\circ} + 9 0^{\circ} n

Solution steps

sin (3 x) cos (x) + cos (3 x) sin (x) = 1

Rewrite using trig identities

sin (3 x) cos (x) + cos (3 x) sin (x)

Use the Angle Sum identity:

sin (s) cos (t) + cos (s) sin (t) = sin (s + t)

= sin (3 x + x)

sin (3 x + x) = 1

General solutions for

sin (3 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}

3 x + x = \frac{π}{2} + 2 πn

3 x + x = \frac{π}{2} + 2 πn

Solve

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

3 x + x = \frac{π}{2} + 2 πn

Add similar elements:

3 x + x = 4 x

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}

2 sin (x) cos (x) + 2 sin (x) - cos (x) - 1 = 0

sin (θ) tan^{2} (θ) - sin (θ) = 0

sin (x) = \frac{4}{5}, sin (18 0^{\circ} - x)

- 11 cos (x) (22) - 11 sin (x) (8) = - 250

3 cos (x) + 1 = 2 sin (x)

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