Trigonometry Examples

$\sin^{2} (x) + 2 \sin (x) - 3 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

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

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

Step 1.2

Factor $u^{2} + 2 u - 3$ using the AC method.

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Step 1.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Step 1.2.2

Write the factored form using these integers.

$(u - 1) (u + 3) = 0$

Step 1.3

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

$(\sin (x) - 1) (\sin (x) + 3) = 0$

Step 2

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

$\sin (x) - 1 = 0$

$\sin (x) + 3 = 0$

Step 3

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

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Step 3.1

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

$\sin (x) - 1 = 0$

Step 3.2

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

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Step 3.2.1

Add $1$ to both sides of the equation.

$\sin (x) = 1$

Step 3.2.2

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

$x = \arcsin (1)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

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

$x = \frac{π}{2}$

Step 3.2.4

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 3.2.5

Simplify $π - \frac{π}{2}$ .

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Step 3.2.5.1

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 3.2.5.2

Combine fractions.

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Step 3.2.5.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 3.2.5.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 3.2.5.3

Simplify the numerator.

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Step 3.2.5.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 3.2.5.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 3.2.6

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

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Step 3.2.6.1

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

$\frac{2 π}{| b |}$

Step 3.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.6.3

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

$\frac{2 π}{1}$

Step 3.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.7

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

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

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Step 4.1

Set $\sin (x) + 3$ equal to $0$ .

$\sin (x) + 3 = 0$

Step 4.2

Solve $\sin (x) + 3 = 0$ for $x$ .

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Step 4.2.1

Subtract $3$ from both sides of the equation.

$\sin (x) = - 3$

Step 4.2.2

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

No solution

Step 5

The final solution is all the values that make $(\sin (x) - 1) (\sin (x) + 3) = 0$ true.

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