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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Factor

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

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

Factor out common term

cos (x)

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

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

Solving each part separately

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

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

cos (x) = 0

General solutions for

cos (x) = 0

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

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

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

No Solution

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

Solve by substitution

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

Let:

sin (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 = sin (x)

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

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

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

No Solution

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

No Solution

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

No Solution

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

No Solution

Combine all the solutions

No Solution

Combine all the solutions

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

Popular Examples

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

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

sec(x)=csc(x)

sec (x) = csc (x)

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

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

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

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

sqrt(3)tan(x)=-1

3 tan (x) = - 1

Frequently Asked Questions (FAQ)

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