Solve over the Interval sin(x)+2sin(x)cos(x)=0 , [0,2pi)

Solution:

Step 1

Extract the common factor $\sin(x)$ from the expression $\sin(x) + 2\sin(x)\cos(x)$.

Step 1.1

Express $\sin(x)$ with an exponent of $1$: $\sin(x) + 2\sin(x)\cos(x) = 0$.

Step 1.2

Extract $\sin(x)$ from $\sin^1(x)$: $\sin(x) \cdot 1 + 2\sin(x)\cos(x) = 0$.

Step 1.3

Remove $\sin(x)$ from $2\sin(x)\cos(x)$: $\sin(x) \cdot 1 + \sin(x)(2\cos(x)) = 0$.

Step 1.4

Factor out $\sin(x)$ completely: $\sin(x)(1 + 2\cos(x)) = 0$.

Step 2

Recognize that if either factor equals zero, the equation is satisfied: $\sin(x) = 0$ or $1 + 2\cos(x) = 0$.

Step 3

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

Step 3.1

Set $\sin(x)$ to zero: $\sin(x) = 0$.

Step 3.2

Find $x$ by solving $\sin(x) = 0$.

Step 3.2.1

Apply the inverse sine function: $x = \arcsin(0)$.

Step 3.2.2

Determine the exact value: $x = 0$.

Step 3.2.3

Find the second solution in the second quadrant: $x = \pi - 0$.

Step 3.2.4

Calculate the difference: $x = \pi$.

Step 3.2.5

Identify the sine function's period.

Step 3.2.5.1

Use the period formula: $\frac{2\pi}{|b|}$.

Step 3.2.5.2

Insert $b = 1$: $\frac{2\pi}{|1|}$.

Step 3.2.5.3

Calculate the absolute value: $\frac{2\pi}{1}$.

Step 3.2.5.4

Divide: $2\pi$.

Step 3.2.6

Express the general solution for sine: $x = 2\pi n$ and $x = \pi + 2\pi n$, where $n$ is any integer.

Step 4

Solve for $x$ when $1 + 2\cos(x) = 0$.

Step 4.1

Set the expression to zero: $1 + 2\cos(x) = 0$.

Step 4.2

Isolate $x$ by solving $1 + 2\cos(x) = 0$.

Step 4.2.1

Subtract $1$ from both sides: $2\cos(x) = -1$.

Step 4.2.2

Divide the equation by $2$.

Step 4.2.2.1

Divide each term: $\frac{2\cos(x)}{2} = \frac{-1}{2}$.

Step 4.2.2.2

Cancel the $2$s: $\cos(x) = \frac{-1}{2}$.

Step 4.2.2.3

Simplify the fraction: $\cos(x) = -\frac{1}{2}$.

Step 4.2.3

Apply the inverse cosine function: $x = \arccos(-\frac{1}{2})$.

Step 4.2.4

Find the exact value: $x = \frac{2\pi}{3}$.

Step 4.2.5

Locate the second solution in the third quadrant: $x = 2\pi - \frac{2\pi}{3}$.

Step 4.2.6

Simplify the expression.

Step 4.2.6.1

Find a common denominator: $x = 2\pi \cdot \frac{3}{3} - \frac{2\pi}{3}$.

Step 4.2.6.2

Combine the fractions: $x = \frac{6\pi - 2\pi}{3}$.

Step 4.2.6.3

Simplify the numerator: $x = \frac{4\pi}{3}$.

Step 4.2.7

Determine the cosine function's period.

Step 4.2.7.1

Use the period formula: $\frac{2\pi}{|b|}$.

Step 4.2.7.2

Insert $b = 1$: $\frac{2\pi}{|1|}$.

Step 4.2.7.3

Calculate the absolute value: $\frac{2\pi}{1}$.

Step 4.2.7.4

Divide: $2\pi$.

Step 4.2.8

Express the general solution for cosine: $x = \frac{2\pi}{3} + 2\pi n$ and $x = \frac{4\pi}{3} + 2\pi n$, where $n$ is any integer.

Step 5

Combine all solutions that satisfy $\sin(x)(1 + 2\cos(x)) = 0$: $x = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3}$, plus multiples of $2\pi$.

Step 6

Simplify the general solutions involving $\pi$: $x = \pi n, \frac{2\pi}{3} + 2\pi n, \frac{4\pi}{3} + 2\pi n$, where $n$ is any integer.

Step 7

Identify solutions within the interval $[0, 2\pi)$.

Step 7.1

Test $n = 0$: $x = 0$ is within the interval.

Step 7.2

Test $n = 0$ for $\frac{2\pi}{3}$: $x = \frac{2\pi}{3}$ is within the interval.

Step 7.3

Test $n = 1$ for $\pi$: $x = \pi$ is within the interval.

Step 7.4

Test $n = 0$ for $\frac{4\pi}{3}$: $x = \frac{4\pi}{3}$ is within the interval.

The final solutions within the interval $[0, 2\pi)$ are $x = 0, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}$.

Knowledge Notes:

• Trigonometric identities and properties are essential for solving trigonometric equations.

• Factoring is a technique used to simplify expressions and solve equations.

• The zero-product property states that if a product of factors equals zero, at least one of the factors must be zero.

• The inverse trigonometric functions, such as $\arcsin$ and $\arccos$, are used to find the angle whose sine or cosine is a given value.

• Trigonometric functions are periodic, with sine and cosine having a period of $2\pi$.

• Solutions to trigonometric equations can be expressed as general solutions that include all possible angles that satisfy the equation.

• When solving over a specific interval, it is necessary to test which general solutions fall within that interval.