Precalculus Examples

$- 2 \sin (x) = - \sin (x) \cos (x)$

Step 1

Add $\sin (x) \cos (x)$ to both sides of the equation.

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

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

$\sin (x) = 0$

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

Step 4

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

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Step 4.1

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

$\sin (x) = 0$

Step 4.2

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

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Step 4.2.1

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

$x = \arcsin (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

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

$x = 0$

Step 4.2.3

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.

$x = π - 0$

Step 4.2.4

Subtract $0$ from $π$ .

$x = π$

Step 4.2.5

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

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Step 4.2.5.1

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

$\frac{2 π}{| b |}$

Step 4.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 4.2.5.3

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

$\frac{2 π}{1}$

Step 4.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.6

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 $- 2 + \cos (x)$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $- 2 + \cos (x)$ equal to $0$ .

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

Step 5.2

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

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Step 5.2.1

Add $2$ to both sides of the equation.

$\cos (x) = 2$

Step 5.2.2

The range of cosine is $- 1 \leq y \leq 1$ . Since $2$ does not fall in this range, there is no solution.

No solution

Step 6

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

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

Step 7

Consolidate the answers.

$x = π n$ , for any integer $n$