Trigonometry Examples

$2 \tan (x) \sin (x) + 2 \sin (x) = \tan (x) + 1$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $\tan (x)$ from both sides of the equation.

$2 \tan (x) \sin (x) + 2 \sin (x) - \tan (x) = 1$

Step 1.2

Subtract $1$ from both sides of the equation.

$2 \tan (x) \sin (x) + 2 \sin (x) - \tan (x) - 1 = 0$

Step 2

Simplify the left side of the equation.

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Step 2.1

Simplify each term.

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Step 2.1.1

Rewrite $\tan (x)$ in terms of sines and cosines.

$2 (\frac{\sin (x)}{\cos (x)}) \cdot \sin (x) + 2 \sin (x) - \tan (x) - 1 = 0$

Step 2.1.2

Combine $2$ and $\frac{\sin (x)}{\cos (x)}$ .

$\frac{2 \sin (x)}{\cos (x)} \cdot \sin (x) + 2 \sin (x) - \tan (x) - 1 = 0$

Step 2.1.3

Multiply $\frac{2 \sin (x)}{\cos (x)} \sin (x)$ .

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Step 2.1.3.1

Combine $\frac{2 \sin (x)}{\cos (x)}$ and $\sin (x)$ .

$\frac{2 \sin (x) \sin (x)}{\cos (x)} + 2 \sin (x) - \tan (x) - 1 = 0$

Step 2.1.3.2

Raise $\sin (x)$ to the power of $1$ .

$\frac{2 (\sin (x) \sin (x))}{\cos (x)} + 2 \sin (x) - \tan (x) - 1 = 0$

Step 2.1.3.3

Raise $\sin (x)$ to the power of $1$ .

$\frac{2 (\sin (x) \sin (x))}{\cos (x)} + 2 \sin (x) - \tan (x) - 1 = 0$

Step 2.1.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{2 {\sin (x)}^{1 + 1}}{\cos (x)} + 2 \sin (x) - \tan (x) - 1 = 0$

Step 2.1.3.5

Add $1$ and $1$ .

$\frac{2 \sin^{2} (x)}{\cos (x)} + 2 \sin (x) - \tan (x) - 1 = 0$

Step 2.1.4

Rewrite $\tan (x)$ in terms of sines and cosines.

$\frac{2 \sin^{2} (x)}{\cos (x)} + 2 \sin (x) - \frac{\sin (x)}{\cos (x)} - 1 = 0$

Step 2.2

Simplify each term.

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Step 2.2.1

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

$\frac{2 (\sin (x) \sin (x))}{\cos (x)} + 2 \sin (x) - \frac{\sin (x)}{\cos (x)} - 1 = 0$

Step 2.2.2

Separate fractions.

$\frac{2 (\sin (x))}{1} \cdot \frac{\sin (x)}{\cos (x)} + 2 \sin (x) - \frac{\sin (x)}{\cos (x)} - 1 = 0$

Step 2.2.3

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

$\frac{2 (\sin (x))}{1} \cdot \tan (x) + 2 \sin (x) - \frac{\sin (x)}{\cos (x)} - 1 = 0$

Step 2.2.4

Divide $2 (\sin (x))$ by $1$ .

$2 (\sin (x)) \tan (x) + 2 \sin (x) - \frac{\sin (x)}{\cos (x)} - 1 = 0$

Step 2.2.5

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

$2 \sin (x) \tan (x) + 2 \sin (x) - \tan (x) - 1 = 0$

Step 3

Factor $2 \sin (x) \tan (x) + 2 \sin (x) - \tan (x) - 1$ .

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Step 3.1

Factor out the greatest common factor from each group.

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Step 3.1.1

Group the first two terms and the last two terms.

$2 \sin (x) \tan (x) + 2 \sin (x) - \tan (x) - 1 = 0$

Step 3.1.2

Factor out the greatest common factor (GCF) from each group.

$2 \sin (x) (\tan (x) + 1) - (\tan (x) + 1) = 0$

Step 3.2

Factor the polynomial by factoring out the greatest common factor, $\tan (x) + 1$ .

$(\tan (x) + 1) (2 \sin (x) - 1) = 0$

Step 4

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

$\tan (x) + 1 = 0$

$2 \sin (x) - 1 = 0$

Step 5

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

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Step 5.1

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

$\tan (x) + 1 = 0$

Step 5.2

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

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Step 5.2.1

Subtract $1$ from both sides of the equation.

$\tan (x) = - 1$

Step 5.2.2

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

$x = \arctan (- 1)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

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

$x = - \frac{π}{4}$

Step 5.2.4

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 5.2.5

Simplify the expression to find the second solution.

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Step 5.2.5.1

Add $2 π$ to $- \frac{π}{4} - π$ .

$x = - \frac{π}{4} - π + 2 π$

Step 5.2.5.2

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

$x = \frac{3 π}{4}$

Step 5.2.6

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

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Step 5.2.6.1

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

$\frac{π}{| b |}$

Step 5.2.6.2

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

$\frac{π}{| 1 |}$

Step 5.2.6.3

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

$\frac{π}{1}$

Step 5.2.6.4

Divide $π$ by $1$ .

$π$

Step 5.2.7

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

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Step 5.2.7.1

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

$- \frac{π}{4} + π$

Step 5.2.7.2

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

$π \cdot \frac{4}{4} - \frac{π}{4}$

Step 5.2.7.3

Combine fractions.

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Step 5.2.7.3.1

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Step 5.2.7.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Step 5.2.7.4

Simplify the numerator.

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Step 5.2.7.4.1

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{4}$

Step 5.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{4}$

Step 5.2.7.5

List the new angles.

$x = \frac{3 π}{4}$

Step 5.2.8

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

Set $2 \sin (x) - 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $2 \sin (x) - 1$ equal to $0$ .

$2 \sin (x) - 1 = 0$

Step 6.2

Solve $2 \sin (x) - 1 = 0$ for $x$ .

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Step 6.2.1

Add $1$ to both sides of the equation.

$2 \sin (x) = 1$

Step 6.2.2

Divide each term in $2 \sin (x) = 1$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 \sin (x) = 1$ by $2$ .

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

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

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

Step 6.2.2.2.1.2

Divide $\sin (x)$ by $1$ .

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

Step 6.2.3

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

$x = \arcsin (\frac{1}{2})$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 6.2.5

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.

$x = π - \frac{π}{6}$

Step 6.2.6

Simplify $π - \frac{π}{6}$ .

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Step 6.2.6.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 6.2.6.2

Combine fractions.

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Step 6.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 6.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 6.2.6.3

Simplify the numerator.

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Step 6.2.6.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 6.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 6.2.7

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

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Step 6.2.7.1

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

$\frac{2 π}{| b |}$

Step 6.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.7.3

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

$\frac{2 π}{1}$

Step 6.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.8

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

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 7

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

$x = \frac{3 π}{4} + π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$