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

The general solution for sqrt(2)sin(x)+sin(2x)=0 is x=2pin,x=pi+2pin,x=(3pi)/4+2pin,x=(5pi)/4+2pin

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Popular >

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

sin (2 x) + sin (x) 2

Use the Double Angle identity:

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

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

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

Factor

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

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

Factor out common term

sin (x)

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

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

Solving each part separately

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

2 + 2 cos (x) = 0 : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

2 + 2 cos (x) = 0

Move

2

to the right side

2 + 2 cos (x) = 0

Subtract

2

from both sides

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

Simplify

2 cos (x) = - 2

2 cos (x) = - 2

Divide both sides by

2

2 cos (x) = - 2

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{- 2}{2}

Simplify

cos (x) = - \frac{2}{2}

cos (x) = - \frac{2}{2}

General solutions for

cos (x) = - \frac{2}{2}

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 = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

Popular Examples

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

2 sin^{2} (x) + sin (x) = 0

tan^2(3x)=3

tan^{2} (3 x) = 3

cos(x)-sqrt(3)sin(x)=2

cos (x) - 3 sin (x) = 2

tan(x)=-3

tan (x) = - 3

sec(θ)=-2

sec (θ) = - 2

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)sin(x)+sin(2x)=0 ?
The general solution for sqrt(2)sin(x)+sin(2x)=0 is x=2pin,x=pi+2pin,x=(3pi)/4+2pin,x=(5pi)/4+2pin