Trigonometry Examples

$2 \sin^{2} (θ) + 3 \sin (θ) = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = \sin (θ)$ . Substitute $u$ for all occurrences of $\sin (θ)$ .

$2 u^{2} + 3 u = 0$

Step 1.2

Factor $u$ out of $2 u^{2} + 3 u$ .

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Step 1.2.1

Factor $u$ out of $2 u^{2}$ .

$u (2 u) + 3 u = 0$

Step 1.2.2

Factor $u$ out of $3 u$ .

$u (2 u) + u \cdot 3 = 0$

Step 1.2.3

Factor $u$ out of $u (2 u) + u \cdot 3$ .

$u (2 u + 3) = 0$

Step 1.3

Replace all occurrences of $u$ with $\sin (θ)$ .

$\sin (θ) (2 \sin (θ) + 3) = 0$

Step 2

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

$\sin (θ) = 0$

$2 \sin (θ) + 3 = 0$

Step 3

Set $\sin (θ)$ equal to $0$ and solve for $θ$ .

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Step 3.1

Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Step 3.2

Solve $\sin (θ) = 0$ for $θ$ .

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Step 3.2.1

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

$θ = \arcsin (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

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

$θ = 0$

Step 3.2.3

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$θ = 180 - 0$

Step 3.2.4

Subtract $0$ from $180$ .

$θ = 180$

Step 3.2.5

Find the period of $\sin (θ)$ .

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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 $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 360 n, 180 + 360 n$ , for any integer $n$

Step 4

Set $2 \sin (θ) + 3$ equal to $0$ and solve for $θ$ .

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Step 4.1

Set $2 \sin (θ) + 3$ equal to $0$ .

$2 \sin (θ) + 3 = 0$

Step 4.2

Solve $2 \sin (θ) + 3 = 0$ for $θ$ .

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Step 4.2.1

Subtract $3$ from both sides of the equation.

$2 \sin (θ) = - 3$

Step 4.2.2

Divide each term in $2 \sin (θ) = - 3$ by $2$ and simplify.

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Step 4.2.2.1

Divide each term in $2 \sin (θ) = - 3$ by $2$ .

$\frac{2 \sin (θ)}{2} = \frac{- 3}{2}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

$\frac{2 \sin (θ)}{2} = \frac{- 3}{2}$

Step 4.2.2.2.1.2

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{- 3}{2}$

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

Move the negative in front of the fraction.

$\sin (θ) = - \frac{3}{2}$

Step 4.2.3

The range of sine is $- 1 \leq y \leq 1$ . Since $- \frac{3}{2}$ does not fall in this range, there is no solution.

No solution

Step 5

The final solution is all the values that make $\sin (θ) (2 \sin (θ) + 3) = 0$ true.

$θ = 360 n, 180 + 360 n$ , for any integer $n$

Step 6

Consolidate the answers.

$θ = 180 n$ , for any integer $n$