Trigonometry Examples

$\sin (2 x) = \cos (3 x)$

Step 1

Subtract $\cos (3 x)$ from both sides of the equation.

$\sin (2 x) - \cos (3 x) = 0$

Step 2

Simplify each term.

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Step 2.1

Apply the sine double-angle identity.

$2 \sin (x) \cos (x) - \cos (3 x) = 0$

Step 2.2

Use the triple-angle identity to transform $\cos (3 x)$ to $4 \cos^{3} (x) - 3 \cos (x)$ .

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

Step 2.3

Apply the distributive property.

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

Step 2.4

Multiply $4$ by $- 1$ .

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

Step 2.5

Multiply $- 3$ by $- 1$ .

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

Step 3

Factor $\cos (x)$ out of $2 \sin (x) \cos (x) - 4 \cos^{3} (x) + 3 \cos (x)$ .

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Step 3.1

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

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

Step 3.2

Factor $\cos (x)$ out of $- 4 \cos^{3} (x)$ .

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

Step 3.3

Factor $\cos (x)$ out of $3 \cos (x)$ .

$\cos (x) (2 \sin (x)) + \cos (x) (- 4 \cos^{2} (x)) + \cos (x) \cdot 3 = 0$

Step 3.4

Factor $\cos (x)$ out of $\cos (x) (2 \sin (x)) + \cos (x) (- 4 \cos^{2} (x))$ .

$\cos (x) (2 \sin (x) - 4 \cos^{2} (x)) + \cos (x) \cdot 3 = 0$

Step 3.5

Factor $\cos (x)$ out of $\cos (x) (2 \sin (x) - 4 \cos^{2} (x)) + \cos (x) \cdot 3$ .

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

Step 4

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

$\cos (x) = 0$

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

Step 5

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

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Step 5.1

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

$\cos (x) = 0$

Step 5.2

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

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Step 5.2.1

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

$x = \arccos (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 5.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{2}$

Step 5.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 5.2.4.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 5.2.4.2

Combine fractions.

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Step 5.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$x = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 5.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 5.2.4.3

Simplify the numerator.

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Step 5.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 5.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 5.2.5

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

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Step 5.2.5.1

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

$\frac{2 π}{| b |}$

Step 5.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.5.3

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

$\frac{2 π}{1}$

Step 5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.6

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

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

Step 6

Set $2 \sin (x) - 4 \cos^{2} (x) + 3$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $2 \sin (x) - 4 \cos^{2} (x) + 3$ equal to $0$ .

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

Step 6.2

Solve $2 \sin (x) - 4 \cos^{2} (x) + 3 = 0$ for $x$ .

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Step 6.2.1

Replace the $- 4 \cos^{2} (x)$ with $- 4 (1 - \sin^{2} (x))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$2 \sin (x) - 4 (1 - \sin^{2} (x)) + 3 = 0$

Step 6.2.2

Simplify each term.

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Step 6.2.2.1

Apply the distributive property.

$2 \sin (x) - 4 \cdot 1 - 4 (- \sin^{2} (x)) + 3 = 0$

Step 6.2.2.2

Multiply $- 4$ by $1$ .

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

Step 6.2.2.3

Multiply $- 1$ by $- 4$ .

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

Step 6.2.3

Add $- 4$ and $3$ .

$2 \sin (x) + 4 \sin^{2} (x) - 1 = 0$

Step 6.2.4

Reorder the polynomial.

$4 \sin^{2} (x) + 2 \sin (x) - 1 = 0$

Step 6.2.5

Substitute $u$ for $\sin (x)$ .

$4 {(u)}^{2} + 2 (u) - 1 = 0$

Step 6.2.6

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6.2.7

Substitute the values $a = 4$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $u$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (4 \cdot - 1)}}{2 \cdot 4}$

Step 6.2.8

Simplify.

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Step 6.2.8.1

Simplify the numerator.

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Step 6.2.8.1.1

Raise $2$ to the power of $2$ .

$u = \frac{- 2 \pm \sqrt{4 - 4 \cdot 4 \cdot - 1}}{2 \cdot 4}$

Step 6.2.8.1.2

Multiply $- 4 \cdot 4 \cdot - 1$ .

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Step 6.2.8.1.2.1

Multiply $- 4$ by $4$ .

$u = \frac{- 2 \pm \sqrt{4 - 16 \cdot - 1}}{2 \cdot 4}$

Step 6.2.8.1.2.2

Multiply $- 16$ by $- 1$ .

$u = \frac{- 2 \pm \sqrt{4 + 16}}{2 \cdot 4}$

Step 6.2.8.1.3

Add $4$ and $16$ .

$u = \frac{- 2 \pm \sqrt{20}}{2 \cdot 4}$

Step 6.2.8.1.4

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 6.2.8.1.4.1

Factor $4$ out of $20$ .

$u = \frac{- 2 \pm \sqrt{4 (5)}}{2 \cdot 4}$

Step 6.2.8.1.4.2

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

$u = \frac{- 2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 4}$

Step 6.2.8.1.5

Pull terms out from under the radical.

$u = \frac{- 2 \pm 2 \sqrt{5}}{2 \cdot 4}$

Step 6.2.8.2

Multiply $2$ by $4$ .

$u = \frac{- 2 \pm 2 \sqrt{5}}{8}$

Step 6.2.8.3

Simplify $\frac{- 2 \pm 2 \sqrt{5}}{8}$ .

$u = \frac{- 1 \pm \sqrt{5}}{4}$

Step 6.2.9

The final answer is the combination of both solutions.

$u = - \frac{1 - \sqrt{5}}{4}, - \frac{1 + \sqrt{5}}{4}$

Step 6.2.10

Substitute $\sin (x)$ for $u$ .

$\sin (x) = - \frac{1 - \sqrt{5}}{4}, - \frac{1 + \sqrt{5}}{4}$

Step 6.2.11

Set up each of the solutions to solve for $x$ .

$\sin (x) = - \frac{1 - \sqrt{5}}{4}$

$\sin (x) = - \frac{1 + \sqrt{5}}{4}$

Step 6.2.12

Solve for $x$ in $\sin (x) = - \frac{1 - \sqrt{5}}{4}$ .

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Step 6.2.12.1

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

$x = \arcsin (- \frac{1 - \sqrt{5}}{4})$

Step 6.2.12.2

Simplify the right side.

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Step 6.2.12.2.1

Evaluate $\arcsin (- \frac{1 - \sqrt{5}}{4})$ .

$x = \frac{π}{10}$

Step 6.2.12.3

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π - \frac{π}{10} + π$

Step 6.2.12.4

Simplify the expression to find the second solution.

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Step 6.2.12.4.1

Subtract $2 π$ from $2 π - \frac{π}{10} + π$ .

$x = 2 π - \frac{π}{10} + π - 2 π$

Step 6.2.12.4.2

The resulting angle of $\frac{9 π}{10}$ is positive, less than $2 π$ , and coterminal with $2 π - \frac{π}{10} + π$ .

$x = \frac{9 π}{10}$

Step 6.2.12.5

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

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Step 6.2.12.5.1

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

$\frac{2 π}{| b |}$

Step 6.2.12.5.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.12.5.3

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

$\frac{2 π}{1}$

Step 6.2.12.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.12.6

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

$x = \frac{π}{10} + 2 π n, \frac{9 π}{10} + 2 π n$ , for any integer $n$

Step 6.2.13

Solve for $x$ in $\sin (x) = - \frac{1 + \sqrt{5}}{4}$ .

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Step 6.2.13.1

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

$x = \arcsin (- \frac{1 + \sqrt{5}}{4})$

Step 6.2.13.2

Simplify the right side.

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Step 6.2.13.2.1

Evaluate $\arcsin (- \frac{1 + \sqrt{5}}{4})$ .

$x = - \frac{3 π}{10}$

Step 6.2.13.3

$x = 2 π + \frac{3 π}{10} + π$

Step 6.2.13.4

Simplify the expression to find the second solution.

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Step 6.2.13.4.1

Subtract $2 π$ from $2 π + \frac{3 π}{10} + π$ .

$x = 2 π + \frac{3 π}{10} + π - 2 π$

Step 6.2.13.4.2

The resulting angle of $\frac{13 π}{10}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{3 π}{10} + π$ .

$x = \frac{13 π}{10}$

Step 6.2.13.5

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

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Step 6.2.13.5.1

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

$\frac{2 π}{| b |}$

Step 6.2.13.5.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.13.5.3

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

$\frac{2 π}{1}$

Step 6.2.13.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.13.6

Add $2 π$ to every negative angle to get positive angles.

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Step 6.2.13.6.1

Add $2 π$ to $- \frac{3 π}{10}$ to find the positive angle.

$- \frac{3 π}{10} + 2 π$

Step 6.2.13.6.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$2 π \cdot \frac{10}{10} - \frac{3 π}{10}$

Step 6.2.13.6.3

Combine fractions.

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Step 6.2.13.6.3.1

Combine $2 π$ and $\frac{10}{10}$ .

$\frac{2 π \cdot 10}{10} - \frac{3 π}{10}$

Step 6.2.13.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 10 - 3 π}{10}$

Step 6.2.13.6.4

Simplify the numerator.

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Step 6.2.13.6.4.1

Multiply $10$ by $2$ .

$\frac{20 π - 3 π}{10}$

Step 6.2.13.6.4.2

Subtract $3 π$ from $20 π$ .

$\frac{17 π}{10}$

Step 6.2.13.6.5

List the new angles.

$x = \frac{17 π}{10}$

Step 6.2.13.7

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

$x = \frac{13 π}{10} + 2 π n, \frac{17 π}{10} + 2 π n$ , for any integer $n$

Step 6.2.14

List all of the solutions.

$x = \frac{π}{10} + 2 π n, \frac{9 π}{10} + 2 π n, \frac{13 π}{10} + 2 π n, \frac{17 π}{10} + 2 π n$ , for any integer $n$

Step 7

The final solution is all the values that make $\cos (x) (2 \sin (x) - 4 \cos^{2} (x) + 3) = 0$ true.

$x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{π}{10} + 2 π n, \frac{9 π}{10} + 2 π n, \frac{13 π}{10} + 2 π n, \frac{17 π}{10} + 2 π n$ , for any integer $n$

Step 8

Consolidate $\frac{π}{2} + 2 π n$ and $\frac{3 π}{2} + 2 π n$ to $\frac{π}{2} + π n$ .

$x = \frac{π}{2} + π n, \frac{π}{10} + 2 π n, \frac{9 π}{10} + 2 π n, \frac{13 π}{10} + 2 π n, \frac{17 π}{10} + 2 π n$ , for any integer $n$