Trigonometry Examples

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

Step 1

Set the denominator in $\frac{2 \cos^{2} (x) - 2 \cos (x) \cos (2 x) - 1}{\sqrt{\sin (x)}}$ equal to $0$ to find where the expression is undefined.

$\sqrt{\sin (x)} = 0$

Step 2

Solve for $x$ .

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Step 2.1

To remove the radical on the left side of the equation, square both sides of the equation.

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

Step 2.2

Simplify each side of the equation.

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Step 2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{\sin (x)}$ as ${\sin (x)}^{\frac{1}{2}}$ .

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Simplify ${({\sin (x)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.2.1.1

Multiply the exponents in ${({\sin (x)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${\sin (x)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 2.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.2.1.1.2.1

Cancel the common factor.

${\sin (x)}^{\frac{1}{2} \cdot 2} = 0^{2}$

Step 2.2.2.1.1.2.2

Rewrite the expression.

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

Step 2.2.2.1.2

Simplify.

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Raising $0$ to any positive power yields $0$ .

$\sin (x) = 0$

Step 2.3

Solve for $x$ .

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Step 2.3.1

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

$x = \arcsin (0)$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

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

$x = 0$

Step 2.3.3

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 2.3.4

Subtract $0$ from $π$ .

$x = π$

Step 2.3.5

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

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Step 2.3.5.1

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

$\frac{2 π}{| b |}$

Step 2.3.5.2

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

$\frac{2 π}{| 1 |}$

Step 2.3.5.3

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

$\frac{2 π}{1}$

Step 2.3.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.3.6

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 2.4

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 2.5

Verify each of the solutions by substituting them into $\sqrt{\sin (x)} = 0$ and solving.

$x = 2 π n$ , for any integer $n$

Step 3

Set the radicand in $\sqrt{\sin (x)}$ less than $0$ to find where the expression is undefined.

$\sin (x) < 0$

Step 4

Solve for $x$ .

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Step 4.1

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

$x < \arcsin (0)$

Step 4.2

Simplify the right side.

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Step 4.2.1

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

$x < 0$

Step 4.3

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 4.4

Subtract $0$ from $π$ .

$x = π$

Step 4.5

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

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Step 4.5.1

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

$\frac{2 π}{| b |}$

Step 4.5.2

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

$\frac{2 π}{| 1 |}$

Step 4.5.3

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

$\frac{2 π}{1}$

Step 4.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.6

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 4.7

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 4.8

Use each root to create test intervals.

$0 < x < π$

$π < x < 2 π$

Step 4.9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 4.9.1

Test a value on the interval $0 < x < π$ to see if it makes the inequality true.

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Step 4.9.1.1

Choose a value on the interval $0 < x < π$ and see if this value makes the original inequality true.

$x = 2$

Step 4.9.1.2

Replace $x$ with $2$ in the original inequality.

$\sin (2) < 0$

Step 4.9.1.3

The left side $0.90929742$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 4.9.2

Test a value on the interval $π < x < 2 π$ to see if it makes the inequality true.

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Step 4.9.2.1

Choose a value on the interval $π < x < 2 π$ and see if this value makes the original inequality true.

$x = 5$

Step 4.9.2.2

Replace $x$ with $5$ in the original inequality.

$\sin (5) < 0$

Step 4.9.2.3

The left side $- 0.95892427$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 4.9.3

Compare the intervals to determine which ones satisfy the original inequality.

$0 < x < π$ False

$π < x < 2 π$ True

$0 < x < π$ False

$π < x < 2 π$ True

Step 4.10

The solution consists of all of the true intervals.

$π + 2 π n < x < 2 π + 2 π n$ , for any integer $n$

Step 5

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

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

Step 6