Trigonometry Examples

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

Step 1

Replace the $\sin^{2} (x)$ with $1 - \cos^{2} (x)$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$(1 - \cos^{2} (x)) + 2 \cos (x) = - 2$

Step 2

Reorder the polynomial.

$- \cos^{2} (x) + 2 \cos (x) + 1 = - 2$

Step 3

Substitute $u$ for $\cos (x)$ .

$- {(u)}^{2} + 2 (u) + 1 = - 2$

Step 4

Add $2$ to both sides of the equation.

$- u^{2} + 2 u + 1 + 2 = 0$

Step 5

Add $1$ and $2$ .

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

Step 6

Factor the left side of the equation.

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Step 6.1

Factor $- 1$ out of $- u^{2} + 2 u + 3$ .

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Step 6.1.1

Factor $- 1$ out of $- u^{2}$ .

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

Step 6.1.2

Factor $- 1$ out of $2 u$ .

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

Step 6.1.3

Rewrite $3$ as $- 1 (- 3)$ .

$- u^{2} - (- 2 u) - 1 \cdot - 3 = 0$

Step 6.1.4

Factor $- 1$ out of $- u^{2} - (- 2 u)$ .

$- (u^{2} - 2 u) - 1 \cdot - 3 = 0$

Step 6.1.5

Factor $- 1$ out of $- (u^{2} - 2 u) - 1 (- 3)$ .

$- (u^{2} - 2 u - 3) = 0$

Step 6.2

Factor.

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Step 6.2.1

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

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Step 6.2.1.1

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 $- 2$ .

$- 3, 1$

Step 6.2.1.2

Write the factored form using these integers.

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

Step 6.2.2

Remove unnecessary parentheses.

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

Step 7

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

$u - 3 = 0$

$u + 1 = 0$

Step 8

Set $u - 3$ equal to $0$ and solve for $u$ .

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Step 8.1

Set $u - 3$ equal to $0$ .

$u - 3 = 0$

Step 8.2

Add $3$ to both sides of the equation.

$u = 3$

Step 9

Set $u + 1$ equal to $0$ and solve for $u$ .

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Step 9.1

Set $u + 1$ equal to $0$ .

$u + 1 = 0$

Step 9.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 10

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

$u = 3, - 1$

Step 11

Substitute $\cos (x)$ for $u$ .

$\cos (x) = 3, - 1$

Step 12

Set up each of the solutions to solve for $x$ .

$\cos (x) = 3$

$\cos (x) = - 1$

Step 13

Solve for $x$ in $\cos (x) = 3$ .

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Step 13.1

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

No solution

Step 14

Solve for $x$ in $\cos (x) = - 1$ .

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Step 14.1

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

$x = \arccos (- 1)$

Step 14.2

Simplify the right side.

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Step 14.2.1

The exact value of $\arccos (- 1)$ is $π$ .

$x = π$

Step 14.3

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - π$

Step 14.4

Subtract $π$ from $2 π$ .

$x = π$

Step 14.5

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

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Step 14.5.1

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

$\frac{2 π}{| b |}$

Step 14.5.2

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

$\frac{2 π}{| 1 |}$

Step 14.5.3

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

$\frac{2 π}{1}$

Step 14.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 14.6

The period of the $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = π + 2 π n$ , for any integer $n$

Step 15

List all of the solutions.

$x = π + 2 π n$ , for any integer $n$