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

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

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Popular >

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

Solution

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

Solution

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

cos (x) \cdot 2 sin (x) cos (x) = 2 cos^{2} (x) sin (x)

cos (x) \cdot 2 sin (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

= 2 sin (x) cos^{2} (x)

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

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

Factor

sin (x) + 2 cos^{2} (x) sin (x) : sin (x) (2 cos^{2} (x) + 1)

sin (x) + 2 cos^{2} (x) sin (x)

Factor out common term

sin (x)

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

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

Solving each part separately

sin (x) = 0 or 2 cos^{2} (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

2 cos^{2} (x) + 1 = 0 :

No Solution

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

Solve by substitution

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

Let:

cos (x) = u

2 u^{2} + 1 = 0

2 u^{2} + 1 = 0 : u = i \frac{1}{2}, u = - i \frac{1}{2}

2 u^{2} + 1 = 0

Move

1

to the right side

2 u^{2} + 1 = 0

Subtract

1

from both sides

2 u^{2} + 1 - 1 = 0 - 1

Simplify

2 u^{2} = - 1

2 u^{2} = - 1

Divide both sides by

2

2 u^{2} = - 1

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{- 1}{2}

Simplify

u^{2} = - \frac{1}{2}

u^{2} = - \frac{1}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - \frac{1}{2}, u = - - \frac{1}{2}

Simplify

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

Apply radical rule:

- a = - 1 a

- \frac{1}{2} = - 1 \frac{1}{2}

= - 1 \frac{1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{2}

Simplify

- - \frac{1}{2} : - i \frac{1}{2}

- - \frac{1}{2}

Simplify

- \frac{1}{2} : i \frac{1}{2}

- \frac{1}{2}

Apply radical rule:

- a = - 1 a

- \frac{1}{2} = - 1 \frac{1}{2}

= - 1 \frac{1}{2}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{2}

= - i \frac{1}{2}

u = i \frac{1}{2}, u = - i \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = i \frac{1}{2} :

No Solution

cos (x) = i \frac{1}{2}

No Solution

cos (x) = - i \frac{1}{2} :

No Solution

cos (x) = - i \frac{1}{2}

No Solution

Combine all the solutions

No Solution

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

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

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

-sin(θ)+2sin^2(θ)=0

- sin (θ) + 2 sin^{2} (θ) = 0

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

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

2tan(x)-3sin(x)=0

2 tan (x) - 3 sin (x) = 0

sin(x)+5=4

sin (x) + 5 = 4

Frequently Asked Questions (FAQ)

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