Trigonometry Examples

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

Step 1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 2

Subtract $1$ from both sides of the equation.

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

Step 3

Simplify the left side.

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Step 3.1

Add $- 2 \sin^{2} (x)$ and $\sin^{2} (x)$ .

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

Step 4

Solve the equation for $x$ .

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Step 4.1

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

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Step 4.1.1

Divide each term in $- \sin^{2} (x) = - 1$ by $- 1$ .

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

Step 4.1.2

Simplify the left side.

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Step 4.1.2.1

Dividing two negative values results in a positive value.

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

Step 4.1.2.2

Divide $\sin^{2} (x)$ by $1$ .

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

Step 4.1.3

Simplify the right side.

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Step 4.1.3.1

Divide $- 1$ by $- 1$ .

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

Step 4.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sin (x) = \pm \sqrt{1}$

Step 4.3

Any root of $1$ is $1$ .

$\sin (x) = \pm 1$

Step 4.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.4.1

First, use the positive value of the $\pm$ to find the first solution.

$\sin (x) = 1$

Step 4.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$\sin (x) = - 1$

Step 4.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$\sin (x) = 1, - 1$

Step 4.5

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

$\sin (x) = 1$

$\sin (x) = - 1$

Step 4.6

Solve for $x$ in $\sin (x) = 1$ .

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Step 4.6.1

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

$x = \arcsin (1)$

Step 4.6.2

Simplify the right side.

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Step 4.6.2.1

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

$x = \frac{π}{2}$

Step 4.6.3

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.

$x = π - \frac{π}{2}$

Step 4.6.4

Simplify $π - \frac{π}{2}$ .

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Step 4.6.4.1

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

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

Step 4.6.4.2

Combine fractions.

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Step 4.6.4.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 4.6.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 4.6.4.3

Simplify the numerator.

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Step 4.6.4.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 4.6.4.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 4.6.5

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

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Step 4.6.5.1

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

$\frac{2 π}{| b |}$

Step 4.6.5.2

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

$\frac{2 π}{| 1 |}$

Step 4.6.5.3

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

$\frac{2 π}{1}$

Step 4.6.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.6.6

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

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

Step 4.7

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

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Step 4.7.1

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

$x = \arcsin (- 1)$

Step 4.7.2

Simplify the right side.

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Step 4.7.2.1

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

$x = - \frac{π}{2}$

Step 4.7.3

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.

$x = 2 π + \frac{π}{2} + π$

Step 4.7.4

Simplify the expression to find the second solution.

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Step 4.7.4.1

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

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 4.7.4.2

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

$x = \frac{3 π}{2}$

Step 4.7.5

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

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Step 4.7.5.1

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

$\frac{2 π}{| b |}$

Step 4.7.5.2

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

$\frac{2 π}{| 1 |}$

Step 4.7.5.3

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

$\frac{2 π}{1}$

Step 4.7.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.7.6

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

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Step 4.7.6.1

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

$- \frac{π}{2} + 2 π$

Step 4.7.6.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 4.7.6.3

Combine fractions.

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Step 4.7.6.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 4.7.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 4.7.6.4

Simplify the numerator.

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Step 4.7.6.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 4.7.6.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 4.7.6.5

List the new angles.

$x = \frac{3 π}{2}$

Step 4.7.7

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

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

Step 4.8

List all of the solutions.

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

Step 4.9

Consolidate the answers.

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