What is the general solution for sin(2pix)cos(pix)+cos(2pix)sin(pix)=-1 ?

The general solution for sin(2pix)cos(pix)+cos(2pix)sin(pix)=-1 is x= 1/2+(2n)/3

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sin(2pix)cos(pix)+cos(2pix)sin(pix)=-1

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

x = \frac{1}{2} + \frac{2 n}{3}

Degrees

x = 28.64788 \dots^{\circ} + 38.19718 \dots^{\circ} n

Solution steps

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

Rewrite using trig identities

sin (2 π x) cos (π x) + cos (2 π x) sin (π x)

Use the Angle Sum identity:

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

= sin (2 π x + π x)

sin (2 π x + π x) = - 1

General solutions for

sin (2 π x + π 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}

2 π x + π x = \frac{3 π}{2} + 2 πn

2 π x + π x = \frac{3 π}{2} + 2 πn

Solve

2 π x + π x = \frac{3 π}{2} + 2 πn : x = \frac{1}{2} + \frac{2 n}{3}

2 π x + π x = \frac{3 π}{2} + 2 πn

Add similar elements:

2 π x + π x = 3 π x

3 π x = \frac{3 π}{2} + 2 πn

Divide both sides by

3 π

3 π x = \frac{3 π}{2} + 2 πn

Divide both sides by

3 π

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

Simplify

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

Simplify

\frac{3 π x}{3 π} : x

\frac{3 π x}{3 π}

Divide the numbers:

\frac{3}{3} = 1

= \frac{π x}{π}

Cancel the common factor:

π

= x

Simplify

\frac{\frac{3 π}{2}}{3 π} + \frac{2 πn}{3 π} : \frac{1}{2} + \frac{2 n}{3}

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

\frac{\frac{3 π}{2}}{3 π} = \frac{1}{2}

\frac{\frac{3 π}{2}}{3 π}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 3 π}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6 π}

Cancel the common factor:

3

= \frac{π}{2 π}

Cancel the common factor:

π

= \frac{1}{2}

\frac{2 πn}{3 π} = \frac{2 n}{3}

\frac{2 πn}{3 π}

Cancel the common factor:

π

= \frac{2 n}{3}

= \frac{1}{2} + \frac{2 n}{3}

x = \frac{1}{2} + \frac{2 n}{3}

x = \frac{1}{2} + \frac{2 n}{3}

x = \frac{1}{2} + \frac{2 n}{3}

x = \frac{1}{2} + \frac{2 n}{3}

0 = - (sin (6.9256 - x) + (sin (- x)))

- 4 π^{2} sin (2 π x) = 0

- 9 sin (y) = 0

6 + 4 cos (2 x) = 2 - 4 cos (x)

x, cos (x) = y

What is the general solution for sin(2pix)cos(pix)+cos(2pix)sin(pix)=-1 ?
The general solution for sin(2pix)cos(pix)+cos(2pix)sin(pix)=-1 is x= 1/2+(2n)/3