Trigonometry Examples

$2 \sin^{2} (x) + \sin (2 x) = 0$

Step 1

Apply the sine double-angle identity.

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

Step 2

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

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Step 2.1

Factor $2 \sin (x)$ out of $2 \sin^{2} (x)$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

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

$\sin (x) = 0$

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

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

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Step 5.1

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

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

Step 5.2

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

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Step 5.2.1

Divide each term in the equation by $\cos (x)$ .

$\frac{\sin (x)}{\cos (x)} + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 5.2.2

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$\tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 5.2.3

Cancel the common factor of $\cos (x)$ .

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Step 5.2.3.1

Cancel the common factor.

$\tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 5.2.3.2

Rewrite the expression.

$\tan (x) + 1 = \frac{0}{\cos (x)}$

Step 5.2.4

Separate fractions.

$\tan (x) + 1 = \frac{0}{1} \cdot \frac{1}{\cos (x)}$

Step 5.2.5

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$\tan (x) + 1 = \frac{0}{1} \cdot \sec (x)$

Step 5.2.6

Divide $0$ by $1$ .

$\tan (x) + 1 = 0 \sec (x)$

Step 5.2.7

Multiply $0$ by $\sec (x)$ .

$\tan (x) + 1 = 0$

Step 5.2.8

Subtract $1$ from both sides of the equation.

$\tan (x) = - 1$

Step 5.2.9

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

$x = \arctan (- 1)$

Step 5.2.10

Simplify the right side.

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Step 5.2.10.1

The exact value of $\arctan (- 1)$ is $- \frac{π}{4}$ .

$x = - \frac{π}{4}$

Step 5.2.11

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

$x = - \frac{π}{4} - π$

Step 5.2.12

Simplify the expression to find the second solution.

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Step 5.2.12.1

Add $2 π$ to $- \frac{π}{4} - π$ .

$x = - \frac{π}{4} - π + 2 π$

Step 5.2.12.2

The resulting angle of $\frac{3 π}{4}$ is positive and coterminal with $- \frac{π}{4} - π$ .

$x = \frac{3 π}{4}$

Step 5.2.13

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

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Step 5.2.13.1

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

$\frac{π}{| b |}$

Step 5.2.13.2

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

$\frac{π}{| 1 |}$

Step 5.2.13.3

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

$\frac{π}{1}$

Step 5.2.13.4

Divide $π$ by $1$ .

$π$

Step 5.2.14

Add $π$ to every negative angle to get positive angles.

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Step 5.2.14.1

Add $π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + π$

Step 5.2.14.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$π \cdot \frac{4}{4} - \frac{π}{4}$

Step 5.2.14.3

Combine fractions.

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Step 5.2.14.3.1

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Step 5.2.14.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Step 5.2.14.4

Simplify the numerator.

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Step 5.2.14.4.1

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{4}$

Step 5.2.14.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{4}$

Step 5.2.14.5

List the new angles.

$x = \frac{3 π}{4}$

Step 5.2.15

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

$x = \frac{3 π}{4} + π n, \frac{3 π}{4} + π n$ , for any integer $n$

Step 6

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

$x = 2 π n, π + 2 π n, \frac{3 π}{4} + π n$ , for any integer $n$

Step 7

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

$x = π n, \frac{3 π}{4} + π n$ , for any integer $n$