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

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

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

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

x = \frac{π}{2} + \frac{2 πn}{3}

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Angle Sum identity:

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

= sin (2 x + x)

sin (2 x + x) = - 1

General solutions for

sin (2 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}

2 x + x = \frac{3 π}{2} + 2 πn

2 x + x = \frac{3 π}{2} + 2 πn

Solve

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

2 x + x = \frac{3 π}{2} + 2 πn

Add similar elements:

2 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}

2 cos (x) = 0.2

cos (a) = - sin (a)

\frac{8}{sin ( 4 5 ^{\circ} )} = \frac{42}{sin ( x )}

sin (x) cos (x) = 3

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

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