Find All Complex Solutions 2sin(x)+sin(2x)=0

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

Step 1

Apply the sine double-angle identity.

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

Step 2

Factor $2 \sin (x)$ out of $2 \sin (x) + 2 \sin (x) \cos (x)$ .

Factor $2 \sin (x)$ out of $2 \sin (x)$ .

$2 \sin (x) \cdot 1 + 2 \sin (x) \cos (x) = 0$

Factor $2 \sin (x)$ out of $2 \sin (x) \cos (x)$ .

$2 \sin (x) \cdot 1 + 2 \sin (x) \cos (x) = 0$

Factor $2 \sin (x)$ out of $2 \sin (x) \cdot 1 + 2 \sin (x) \cos (x)$ .

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

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sin (x) = 0$

$1 + \cos (x) = 0$

Step 4

Set $\sin (x)$ equal to $0$ and solve for $x$ .

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Solve $\sin (x) = 0$ for $x$ .

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (0)$

Simplify the right side.

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - 0$

Subtract $0$ from $π$ .

$x = π$

Find the period of $\sin (x)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 5

Set $1 + \cos (x)$ equal to $0$ and solve for $x$ .

Set $1 + \cos (x)$ equal to $0$ .

$1 + \cos (x) = 0$

Solve $1 + \cos (x) = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

Take the inverse cosine of both sides of the equation to extract $x$ from inside the cosine.

$x = \arccos (- 1)$

Simplify the right side.

The exact value of $\arccos (- 1)$ is $π$ .

$x = π$

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - π$

Subtract $π$ from $2 π$ .

$x = π$

Find the period of $\cos (x)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = π + 2 π n$ , for any integer $n$

Step 6

The final solution is all the values that make $2 \sin (x) (1 + \cos (x)) = 0$ true.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 7

Consolidate the answers.

$x = π n$ , for any integer $n$

Find All Complex Solutions 2sin(x)+sin(2x)=0

Related Questions

Support LCMGCF.com by disabling your adblocker.