Trigonometry Examples

$3 \sin^{2} (θ) - 4 = - 4 \sin (θ)$

Step 1

Add $4 \sin (θ)$ to both sides of the equation.

$3 \sin^{2} (θ) - 4 + 4 \sin (θ) = 0$

Step 2

Factor by grouping.

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Step 2.1

Reorder terms.

$3 \sin^{2} (θ) + 4 \sin (θ) - 4 = 0$

Step 2.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 4 = - 12$ and whose sum is $b = 4$ .

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Step 2.2.1

Factor $4$ out of $4 \sin (θ)$ .

$3 \sin^{2} (θ) + 4 \sin (θ) - 4 = 0$

Step 2.2.2

Rewrite $4$ as $- 2$ plus $6$

$3 \sin^{2} (θ) + (- 2 + 6) \sin (θ) - 4 = 0$

Step 2.2.3

Apply the distributive property.

$3 \sin^{2} (θ) - 2 \sin (θ) + 6 \sin (θ) - 4 = 0$

Step 2.3

Factor out the greatest common factor from each group.

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Step 2.3.1

Group the first two terms and the last two terms.

$3 \sin^{2} (θ) - 2 \sin (θ) + 6 \sin (θ) - 4 = 0$

Step 2.3.2

Factor out the greatest common factor (GCF) from each group.

$\sin (θ) (3 \sin (θ) - 2) + 2 (3 \sin (θ) - 2) = 0$

Step 2.4

Factor the polynomial by factoring out the greatest common factor, $3 \sin (θ) - 2$ .

$(3 \sin (θ) - 2) (\sin (θ) + 2) = 0$

Step 3

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

$3 \sin (θ) - 2 = 0$

$\sin (θ) + 2 = 0$

Step 4

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

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Step 4.1

Set $3 \sin (θ) - 2$ equal to $0$ .

$3 \sin (θ) - 2 = 0$

Step 4.2

Solve $3 \sin (θ) - 2 = 0$ for $θ$ .

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Step 4.2.1

Add $2$ to both sides of the equation.

$3 \sin (θ) = 2$

Step 4.2.2

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

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Step 4.2.2.1

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

$\frac{3 \sin (θ)}{3} = \frac{2}{3}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $3$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

$\frac{3 \sin (θ)}{3} = \frac{2}{3}$

Step 4.2.2.2.1.2

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{2}{3}$

Step 4.2.3

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

$θ = \arcsin (\frac{2}{3})$

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

Evaluate $\arcsin (\frac{2}{3})$ .

$θ = 41.81031489$

Step 4.2.5

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.

$θ = 180 - 41.81031489$

Step 4.2.6

Subtract $41.81031489$ from $180$ .

$θ = 138.1896851$

Step 4.2.7

Find the period of $\sin (θ)$ .

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Step 4.2.7.1

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

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

Step 4.2.7.2

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

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

Step 4.2.7.3

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

$\frac{360}{1}$

Step 4.2.7.4

Divide $360$ by $1$ .

$360$

Step 4.2.8

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

$θ = 41.81031489 + 360 n, 138.1896851 + 360 n$ , for any integer $n$

Step 5

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

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Step 5.1

Set $\sin (θ) + 2$ equal to $0$ .

$\sin (θ) + 2 = 0$

Step 5.2

Solve $\sin (θ) + 2 = 0$ for $θ$ .

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Step 5.2.1

Subtract $2$ from both sides of the equation.

$\sin (θ) = - 2$

Step 5.2.2

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

No solution

Step 6

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

$θ = 41.81031489 + 360 n, 138.1896851 + 360 n$ , for any integer $n$