Trigonometry Examples

$9 \sin^{2} (x) \tan (x) = 16 \tan (x)$

Step 1

Subtract $16 \tan (x)$ from both sides of the equation.

$9 \sin^{2} (x) \tan (x) - 16 \tan (x) = 0$

Step 2

Factor $9 \sin^{2} (x) \tan (x) - 16 \tan (x)$ .

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Step 2.1

Factor $\tan (x)$ out of $9 \sin^{2} (x) \tan (x) - 16 \tan (x)$ .

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Step 2.1.1

Factor $\tan (x)$ out of $9 \sin^{2} (x) \tan (x)$ .

$\tan (x) (9 \sin^{2} (x)) - 16 \tan (x) = 0$

Step 2.1.2

Factor $\tan (x)$ out of $- 16 \tan (x)$ .

$\tan (x) (9 \sin^{2} (x)) + \tan (x) \cdot - 16 = 0$

Step 2.1.3

Factor $\tan (x)$ out of $\tan (x) (9 \sin^{2} (x)) + \tan (x) \cdot - 16$ .

$\tan (x) (9 \sin^{2} (x) - 16) = 0$

Step 2.2

Rewrite $9 \sin^{2} (x)$ as ${(3 \sin (x))}^{2}$ .

$\tan (x) ({(3 \sin (x))}^{2} - 16) = 0$

Step 2.3

Rewrite $16$ as $4^{2}$ .

$\tan (x) ({(3 \sin (x))}^{2} - 4^{2}) = 0$

Step 2.4

Factor.

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Step 2.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 \sin (x)$ and $b = 4$ .

$\tan (x) ((3 \sin (x) + 4) (3 \sin (x) - 4)) = 0$

Step 2.4.2

Remove unnecessary parentheses.

$\tan (x) (3 \sin (x) + 4) (3 \sin (x) - 4) = 0$

Step 3

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

$\tan (x) = 0$

$3 \sin (x) + 4 = 0$

$3 \sin (x) - 4 = 0$

Step 4

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

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Step 4.1

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

$\tan (x) = 0$

Step 4.2

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

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Step 4.2.1

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

$x = \arctan (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

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

$x = 0$

Step 4.2.3

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

$x = π + 0$

Step 4.2.4

Add $π$ and $0$ .

$x = π$

Step 4.2.5

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

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Step 4.2.5.1

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

$\frac{π}{| b |}$

Step 4.2.5.2

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

$\frac{π}{| 1 |}$

Step 4.2.5.3

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

$\frac{π}{1}$

Step 4.2.5.4

Divide $π$ by $1$ .

$π$

Step 4.2.6

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

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

Step 5

Set $3 \sin (x) + 4$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $3 \sin (x) + 4$ equal to $0$ .

$3 \sin (x) + 4 = 0$

Step 5.2

Solve $3 \sin (x) + 4 = 0$ for $x$ .

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Step 5.2.1

Subtract $4$ from both sides of the equation.

$3 \sin (x) = - 4$

Step 5.2.2

Divide each term in $3 \sin (x) = - 4$ by $3$ and simplify.

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Step 5.2.2.1

Divide each term in $3 \sin (x) = - 4$ by $3$ .

$\frac{3 \sin (x)}{3} = \frac{- 4}{3}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $3$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{3 \sin (x)}{3} = \frac{- 4}{3}$

Step 5.2.2.2.1.2

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

$\sin (x) = \frac{- 4}{3}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Move the negative in front of the fraction.

$\sin (x) = - \frac{4}{3}$

Step 5.2.3

The range of sine is $- 1 \leq y \leq 1$ . Since $- \frac{4}{3}$ does not fall in this range, there is no solution.

No solution

Step 6

The final solution is all the values that make $\tan (x) (3 \sin (x) + 4) (3 \sin (x) - 4) = 0$ true.

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

Step 7

Consolidate the answers.

$x = π n$ , for any integer $n$