What is the general solution for 2sin(x)+sin(2x)=0 ?

The general solution for 2sin(x)+sin(2x)=0 is x=2pin,x=pi+2pin

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2sin(x)+sin(2x)=0

2 sin (x) + sin (2 x) = 0

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (x) + sin (2 x) = 0

Rewrite using trig identities

sin (2 x) + 2 sin (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

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

Factor

2 sin (x) + 2 cos (x) sin (x) : 2 sin (x) (cos (x) + 1)

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

Factor out common term

2 sin (x)

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

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

Solving each part separately

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

cos (x) = - 1

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

cos (x) = 0.6

tan (2 x) - 1 = 0

tan (x) = - \frac{1}{3}

cot (3 x) = - \frac{3}{3}, 0 \leq x \leq 2 π

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

What is the general solution for 2sin(x)+sin(2x)=0 ?
The general solution for 2sin(x)+sin(2x)=0 is x=2pin,x=pi+2pin