Trigonometry Examples

$\sin (x) \cos (x) + \cos (x) = 0$

Step 1

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

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Step 1.1

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

$\cos (x) \sin (x) + \cos (x) = 0$

Step 1.2

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

$\cos (x) \sin (x) + \cos (x) = 0$

Step 1.3

Factor $\cos (x)$ out of $\cos^{1} (x)$ .

$\cos (x) \sin (x) + \cos (x) \cdot 1 = 0$

Step 1.4

Factor $\cos (x)$ out of $\cos (x) \sin (x) + \cos (x) \cdot 1$ .

$\cos (x) (\sin (x) + 1) = 0$

Step 2

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

$\cos (x) = 0$

$\sin (x) + 1 = 0$

Step 3

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

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Step 3.1

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

$\cos (x) = 0$

Step 3.2

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

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Step 3.2.1

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

$x = \arccos (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

The exact value of $\arccos (0)$ is $90$ .

$x = 90$

Step 3.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$x = 360 - 90$

Step 3.2.4

Subtract $90$ from $360$ .

$x = 270$

Step 3.2.5

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

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Step 3.2.5.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 3.2.5.2

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

$\frac{360}{| 1 |}$

Step 3.2.5.3

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

$\frac{360}{1}$

Step 3.2.5.4

Divide $360$ by $1$ .

$360$

Step 3.2.6

The period of the $\cos (x)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 90 + 360 n, 270 + 360 n$ , for any integer $n$

Step 4

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

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Step 4.1

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

$\sin (x) + 1 = 0$

Step 4.2

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

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 4.2.2

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

$x = \arcsin (- 1)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

The exact value of $\arcsin (- 1)$ is $- 90$ .

$x = - 90$

Step 4.2.4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $360$ , to find a reference angle. Next, add this reference angle to $180$ to find the solution in the third quadrant.

$x = 360 + 90 + 180$

Step 4.2.5

Simplify the expression to find the second solution.

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Step 4.2.5.1

Subtract $360 °$ from $360 + 90 + 180 °$ .

$x = 360 + 90 + 180 ° - 360 °$

Step 4.2.5.2

The resulting angle of $270 °$ is positive, less than $360 °$ , and coterminal with $360 + 90 + 180$ .

$x = 270 °$

Step 4.2.6

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

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Step 4.2.6.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 4.2.6.2

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

$\frac{360}{| 1 |}$

Step 4.2.6.3

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

$\frac{360}{1}$

Step 4.2.6.4

Divide $360$ by $1$ .

$360$

Step 4.2.7

Add $360$ to every negative angle to get positive angles.

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Step 4.2.7.1

Add $360$ to $- 90$ to find the positive angle.

$- 90 + 360$

Step 4.2.7.2

Subtract $90$ from $360$ .

$270$

Step 4.2.7.3

List the new angles.

$x = 270$

Step 4.2.8

The period of the $\sin (x)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 270 + 360 n, 270 + 360 n$ , for any integer $n$

Step 5

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

$x = 90 + 360 n, 270 + 360 n$ , for any integer $n$

Step 6

Consolidate the answers.

$x = 90 + 180 n$ , for any integer $n$